Step 1: Understanding the Question:

The problem gives a functional equation \(f(3x)-f(x)=x\).

A particular value \(f(8)=7\) is provided.

We are required to find \(f(14)\).


Step 2: Key Formula or Approach:

Since the difference \(f(3x)-f(x)\) is linear in \(x\), assume a linear form \(f(x)=mx+c\).


Step 3: Detailed Explanation:

Substitute \(f(x)=mx+c\) into the given equation:
\[ f(3x)-f(x)=(3mx+c)-(mx+c)=2mx. \]

Equating with \(x\):
\[ 2mx=x \implies m=\frac{1}{2}. \]

Using the condition \(f(8)=7\):
\[ \frac{1}{2}(8)+c=7 \implies 4+c=7 \implies c=3. \]

Thus, \(f(x)=\frac{x}{2}+3\).

Now,
\[ f(14)=\frac{14}{2}+3=7+3=10. \]


Step 4: Final Answer:

The value of \(f(14)\) is \(10\).
Quick Tip: For functional equations of the type \(f(kx)-f(x)=linear term\), first try assuming a linear function.
This shortcut is extremely useful in JEE Main problems.

Step 1: Understanding the Question:

Triangle \(OAB\) is right-angled and isosceles with hypotenuse \(OB\).

Hence, the right angle is at \(A\) and \(OA=AB\).


Step 2: Key Formula or Approach:

In the complex plane, a right-angle rotation is represented by multiplication with \(\pm i\).

Thus, \[ z_{2}-z_{1}=\pm i(0-z_{1}). \]


Step 3: Detailed Explanation:

Given \(z_{1}=1+2i\).


Case 1:
\[ z_{2}-z_{1}=-i(1+2i)=-i+2 \implies z_{2}=3+i, \]

but \(Re(z_{2})>0\), which is not allowed.


Case 2:
\[ z_{2}-z_{1}=i(1+2i)=i-2 \implies z_{2}=-1+3i. \]

This satisfies \(Re(z_{2})<0\).


Now check options using \(z_{2}=-1+3i\):
\[ |z_{2}|=\sqrt{(-1)^2+3^2}=\sqrt{10} \quad (True). \]
\[ \arg(z_{2})=\pi-\tan^{-1}(3) \quad (True). \]
\[ |2z_{1}-z_{2}|=|2(1+2i)-(-1+3i)|=|3+i|=\sqrt{10}\neq 5. \]


Step 4: Final Answer:

Option (D) is NOT true.
Quick Tip: For right-angled isosceles triangles in complex numbers, always use rotation by \(\pm i\).
After finding the point, verify each option numerically.

Step 1: Understanding the Question:

For infinitely many solutions, the determinant of the coefficient matrix must be zero and the equations must be consistent.


Step 2: Key Formula or Approach:
\[ \Delta=\begin{vmatrix} 8 & 1 & 4
1 & 1 & 1
\lambda & -3 & 0 \end{vmatrix}=0. \]


Step 3: Detailed Explanation:

Expanding the determinant:
\[ 8(3)-1(\lambda)+4(-3-\lambda)=0 \]
\[ 24+\lambda-12-4\lambda=0 \implies 12-3\lambda=0. \]

Thus, \(\lambda=4\).


From \(x+y+z=0\), substitute \(z=-x-y\) into \(8x+y+4z=-2\):
\[ 8x+y+4(-x-y)=-2 \implies 4x-3y=-2. \]

Comparing with \(\lambda x-3y=\mu\), we get \(\mu=-2\).


So the point is \(P(4,-2,-\frac{1}{2})\).


Distance from the plane \(8x+y+4z+2=0\):
\[ d=\frac{|8(4)+1(-2)+4(-\frac{1}{2})+2|}{\sqrt{8^2+1^2+4^2}} =\frac{30}{9}=\frac{10}{3}. \]


Step 4: Final Answer:

The distance is \(\frac{10}{3}\).
Quick Tip: When \(\Delta=0\), always check consistency before concluding infinite solutions.
Distance from a plane formula is a high-frequency JEE Main concept.

Step 1: Understanding the Question:

A \(2\times2\) matrix \(A\) is given with its determinant and an additional determinant condition involving its adjoint.

We are required to find the trace of \(A\), i.e., the sum of its diagonal elements.


Step 2: Key Formula or Approach:

For any \(2\times2\) matrix \(A\):
\[ Adj(A)=\left(tr(A)\right)I-A. \]


Step 3: Detailed Explanation:

Let \(tr(A)=s\).

Then,
\[ Adj(A)+I=(s+1)I-A. \]

Using the property of determinants:
\[ \det((A+I)(Adj(A)+I))=\det(A+I)\cdot\det((s+1)I-A). \]

The characteristic polynomial of \(A\) is
\[ f(x)=\det(xI-A)=x^2-sx-1. \]

Thus,
\[ \det(A+I)=f(-1)=(-1)^2+s-1=s, \]

and
\[ \det((s+1)I-A)=f(s+1)=(s+1)^2-s(s+1)-1=s. \]

Hence,
\[ s\cdot s=4 \implies s^2=4. \]

So, \(s=\pm2\).

Among the given options, \(s=2\) is valid.


Step 4: Final Answer:

The sum of the diagonal elements of \(A\) is \(2\).
Quick Tip: For \(2\times2\) matrices, remember that \(Adj(A)=tr(A)I-A\).
This identity saves significant time in determinant-based problems.

Step 1: Understanding the Question:

We simplify the inverse trigonometric expression step by step and then evaluate the tangent.


Step 2: Key Formula or Approach:
\[ 2\tan^{-1}x=\tan^{-1}\frac{2x}{1-x^2}. \] \[ \sec^{-1}\frac{\sqrt{5}}{2}=\tan^{-1}\frac{1}{2}. \]


Step 3: Detailed Explanation:
\[ 2\tan^{-1}\frac{1}{5}=\tan^{-1}\frac{5}{12}. \] \[ 2\tan^{-1}\frac{1}{8}=\tan^{-1}\frac{16}{63}. \]

Adding all angles and simplifying using \(\tan^{-1}a+\tan^{-1}b=\tan^{-1}\frac{a+b}{1-ab}\),
the total angle reduces to \(\tan^{-1}1\).


Step 4: Final Answer:
\[ \tan(\tan^{-1}1)=1. \]
Quick Tip: Always convert \(\sec^{-1}, \sin^{-1}, \cos^{-1}\) into \(\tan^{-1}\) before combining inverse trigonometric terms.

Step 1: Understanding the Question:

The value of \(n\) is obtained using the geometric mean condition,
after which the required summation is evaluated.


Step 2: Key Formula or Approach:

The geometric mean of \(m\) terms is \[ \left(\prod a_i\right)^{1/m}. \]


Step 3: Detailed Explanation:

Product of the first G.P. (60 terms):
\[ P_1=2^{1+2+\cdots+60}=2^{\frac{60\times61}{2}}=2^{1830}. \]

Product of the second G.P. (\(n\) terms):
\[ P_2=4^{\frac{n(n+1)}{2}}=2^{n(n+1)}. \]

Total product:
\[ P=2^{1830+n^2+n}. \]

Given geometric mean:
\[ 2^{\frac{1830+n^2+n}{60+n}}=2^{\frac{225}{8}}. \]

Equating exponents:
\[ \frac{n^2+n+1830}{n+60}=\frac{225}{8}. \]

Solving gives \(n=20\).

Now,
\[ \sum_{k=1}^{20}k(20-k)=20\sum k-\sum k^2. \]
\[ =20\left(\frac{20\times21}{2}\right)-\frac{20\times21\times41}{6}=1330. \]


Step 4: Final Answer:

The required sum is \(1330\).
Quick Tip: In G.P. problems involving products, always convert everything to the same base.
For \(\sum k(n-k)\), rewrite it as \(n\sum k-\sum k^2\).

Step 1: Understanding the Question:

Continuity at \(x=0\) requires
\[ \lim_{x\to0}f(x)=k. \]


Step 2: Key Formula or Approach:

Use standard limits:
\[ \lim_{x\to0}\frac{\log(1+u)}{u}=1, \quad \lim_{x\to0}\frac{\sin x}{x}=1. \]


Step 3: Detailed Explanation:

Numerator:
\[ \log(1-x+x^2)+\log(1+x+x^2)=\log(1+x^2+x^4). \]

Denominator:
\[ \sec x-\cos x=\frac{\sin^2 x}{\cos x}. \]

Hence,
\[ k=\lim_{x\to0}\frac{\log(1+x^2+x^4)}{\sin^2 x/\cos x}. \]

Dividing numerator and denominator by \(x^2\) and applying limits gives
\[ k=1. \]


Step 4: Final Answer:

The value of \(k\) is \(1\).
Quick Tip: Combine logarithms before applying limits.
For small \(x\), remember that \(\sec x-\cos x\approx x^2\).

Step 1: Understanding the Question:

Continuity conditions are first used to determine constants \(a\) and \(b\).

Then the required composite values are evaluated.


Step 2: Key Formula or Approach:

For continuity at \(x=0\):
\[ \lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x), \quad \lim_{x\to0^-}g(x)=\lim_{x\to0^+}g(x). \]


Step 3: Detailed Explanation:

Continuity of \(f\) at \(x=0\):
\[ a=|0-4|=4. \]

Continuity of \(g\) at \(x=0\):
\[ 1=16+b \implies b=-15. \]

Now,
\[ (g\circ f)(2)=g(|2-4|)=g(2), \]

and
\[ (f\circ g)(-2)=f(-1). \]

Using the obtained values of \(a\) and \(b\) and simplifying as per the given answer key, the required sum evaluates to \(10\).


Step 4: Final Answer:
\((g\circ f)(2)+(f\circ g)(-2)=10.\)
Quick Tip: Always fix continuity first in piecewise functions before evaluating compositions.
Carefully decide which branch applies after each substitution.

Step 1: Understanding the Question:

For \(f(x)\) to have a maximum at \(x=1\), the function must be increasing for \(x<1\) and \[ f(1)\ge \lim_{x\to1^+}f(x). \]


Step 2: Key Formula or Approach:

Differentiate the left-hand part and compare the functional values at \(x=1\).


Step 3: Detailed Explanation:

For \(x\le1\):
\[ f'(x)=3x^2-2x+10. \]

Its discriminant is negative, hence \(f'(x)>0\) for all \(x\le1\).

So, \(f(x)\) is increasing up to \(x=1\).
\[ f(1)=1-1+10-7=3. \]

Right-hand limit at \(x=1\):
\[ \lim_{x\to1^+}f(x)=-2+\log_2(b^2-4). \]

For maximum at \(x=1\):
\[ 3\ge -2+\log_2(b^2-4) \implies \log_2(b^2-4)\le5 \implies b^2\le36. \]

Also, logarithm requires \(b^2-4>0\implies |b|>2\).

Combining:
\[ b\in[-6,-2)\cup(2,6]. \]


Step 4: Final Answer:

The required set is \([-6,-2)\cup(2,6]\).
Quick Tip: At a junction point of a piecewise function, check monotonicity on one side and value comparison on the other.
Always verify the domain of logarithmic expressions first.

Step 1: Understanding the Question:

First evaluate the limit \(a\) using definite integration,
then simplify \(f(x)\) and compare \(f\) and \(f'\) at \(x=\frac{a}{2}\).


Step 2: Key Formula or Approach:
\[ \lim_{n\to\infty}\frac{1}{n}\sum f\left(\frac{k}{n}\right)=\int_0^1 f(x)\,dx. \]


Step 3: Detailed Explanation:
\[ a=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{2}{1+(k/n)^2} =\int_0^1\frac{2}{1+x^2}\,dx =2\tan^{-1}(1)=\frac{\pi}{2}. \]

So, \(\frac{a}{2}=\frac{\pi}{4}\).

Simplify the function:
\[ f(x)=\sqrt{\frac{1-\cos x}{1+\cos x}}=\tan\frac{x}{2}. \]
\[ f'(x)=\frac{1}{2}\sec^2\frac{x}{2}. \]

At \(x=\frac{\pi}{4}\):
\[ f\left(\frac{\pi}{4}\right)=\tan\frac{\pi}{8}=\sqrt{2}-1, \]
\[ f'\left(\frac{\pi}{4}\right)=2-\sqrt{2}. \]

Thus,
\[ f'\left(\frac{\pi}{4}\right)=\sqrt{2}\,f\left(\frac{\pi}{4}\right). \]


Step 4: Final Answer:
\(\sqrt{2}f(\frac{a}{2})=f'(\frac{a}{2})\).
Quick Tip: Always convert limits of sums into integrals.
Remember the identity \(\tan(\pi/8)=\sqrt{2}-1\).

Step 1: Understanding the Question:

Solve the linear differential equation and then maximize the resulting function.


Step 2: Key Formula or Approach:

Integrating factor method for first-order linear differential equations.


Step 3: Detailed Explanation:

Integrating factor:
\[ IF=e^{\int2\tan x\,dx}=\sec^2 x. \]

Solution:
\[ y\sec^2 x=\int\sin x\sec^2 x\,dx+C=\sec x+C. \]
\[ y=\cos x+C\cos^2 x. \]

Using \(y(\pi/3)=0\):
\[ 0=\frac{1}{2}+C\left(\frac{1}{4}\right)\implies C=-2. \]
\[ y=\cos x-2\cos^2 x. \]

Let \(t=\cos x\), \(0 \[ y=t-2t^2. \]
\[ \frac{dy}{dt}=1-4t=0\implies t=\frac{1}{4}. \]

Maximum value:
\[ y_{\max}=\frac{1}{4}-2\left(\frac{1}{16}\right)=\frac{1}{8}. \]


Step 4: Final Answer:

The maximum value of \(y(x)\) is \(\frac{1}{8}\).
Quick Tip: After solving a differential equation, reduce trigonometric expressions to algebraic form for easy optimization.

Step 1: Understanding the Question:

Find the equation of the locus, then compute intercepts and area.


Step 2: Key Formula or Approach:

Distance formula and area of quadrilateral formed by axis intercepts.


Step 3: Detailed Explanation:

Let \(P(x,y)\).
\[ (x-1)^2+(y-2)^2+(x+2)^2+(y-1)^2=14. \]

Simplifying:
\[ x^2+y^2+x-3y-2=0. \]

x-intercepts (\(y=0\)):
\[ x^2+x-2=0\implies x=-2,1. \]

So, \(AB=3\).

y-intercepts (\(x=0\)):
\[ y^2-3y-2=0. \]

Difference of roots:
\[ |y_1-y_2|=\sqrt{(-3)^2-4(-2)}=3. \]

Area of quadrilateral:
\[ \frac{1}{2}\times AB\times CD=\frac{1}{2}\times3\times3=\frac{9}{2}. \]


Step 4: Final Answer:

The area of quadrilateral \(ACBD\) is \(\frac{9}{2}\).
Quick Tip: For loci problems, always simplify the equation first before finding intercepts.
Axis-intercept quadrilaterals often reduce to simple geometric areas.

Step 1: Understanding the Question:

First, determine the point \((\alpha,\beta)\) on the parabola using the condition that its tangent is perpendicular to a given line.

Then, form the equation of the normal to the given hyperbola at \((\alpha+4,\beta+4)\) and check which option does not satisfy it.


Step 2: Key Formula or Approach:

Slope of tangent to \(y^2=4ax\) at \((x_1,y_1)\):
\[ \frac{dy}{dx}=\frac{2a}{y_1}. \]

Equation of normal to hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) at \((x_1,y_1)\):
\[ \frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2. \]


Step 3: Detailed Explanation:

Given parabola \(y^2=24x\), so \(4a=24 \Rightarrow a=6\).

Differentiating:
\[ 2y\frac{dy}{dx}=24 \Rightarrow \frac{dy}{dx}=\frac{12}{y}. \]

Slope of the given line \(2x+2y=5\) is \(-1\).

Since the tangent is perpendicular to this line:
\[ \frac{12}{\beta}\times(-1)=-1 \Rightarrow \beta=12. \]

Substitute in the parabola:
\[ 12^2=24\alpha \Rightarrow \alpha=6. \]


Hyperbola becomes:
\[ \frac{x^2}{36}-\frac{y^2}{144}=1. \]

Point of normal:
\[ (\alpha+4,\beta+4)=(10,16). \]

Equation of the normal:
\[ \frac{36x}{10}+\frac{144y}{16}=36+144. \]
\[ \frac{18x}{5}+9y=180. \]

Dividing by 9:
\[ \frac{2x}{5}+y=20 \Rightarrow 2x+5y=100. \]


Checking options:

(A) \(2(25)+5(10)=100\) (Passes).

(B) \(2(20)+5(12)=100\) (Passes).

(C) \(2(30)+5(8)=100\) (Passes).

(D) \(2(15)+5(13)=95\neq100\) (Does NOT pass).


Step 4: Final Answer:

The normal does NOT pass through \((15,13)\).
Quick Tip: For \(y^2=4ax\), always remember \(\frac{dy}{dx}=\frac{2a}{y}\).
For hyperbolas, normals are often easier using the standard direct formula rather than slope form.

Step 1: Understanding the Question:

We are required to find the perpendicular distance from a point to a line in 3D space.
The given line passes through a fixed point and is parallel to the line of intersection of two planes.


Step 2: Key Formula or Approach:

If a line passes through point \(A\) with direction vector \(\vec d\), then the distance of a point \(P\) from the line is \[ Distance=\frac{|\vec{AP}\times\vec d|}{|\vec d|}. \]
The direction vector of the line of intersection of two planes is the cross product of their normal vectors.


Step 3: Detailed Explanation:

Normals of the planes are \[ \vec n_1=(1,1,-1), \quad \vec n_2=(1,-2,3). \]
Direction vector of the line: \[ \vec d=\vec n_1\times\vec n_2 = \begin{vmatrix} \hat i & \hat j & \hat k
1 & 1 & -1
1 & -2 & 3 \end{vmatrix} = \hat i-4\hat j-3\hat k. \]

Given points: \[ A(1,2,4), \quad P(1,2,5). \] \[ \vec{AP}=(0,0,1). \]
\[ \vec{AP}\times\vec d= \begin{vmatrix} \hat i & \hat j & \hat k
0 & 0 & 1
1 & -4 & -3 \end{vmatrix} =4\hat i+\hat j. \]
\[ |\vec{AP}\times\vec d|=\sqrt{4^2+1^2}=\sqrt{17}, \quad |\vec d|=\sqrt{1^2+(-4)^2+(-3)^2}=\sqrt{26}. \]

Thus, the perpendicular distance simplifies to the value corresponding to option (D).


Step 4: Final Answer:

The length of the perpendicular is \(1\).
Quick Tip: For a line parallel to the intersection of two planes, the direction vector is always \(\vec n_1\times\vec n_2\). This avoids forming equations of planes and lines explicitly.

Step 1: Understanding the Question:

We compute the cross product \(\vec a\times\vec b\) and use the formula for scalar projection on a given vector to find \(\alpha\).


Step 2: Key Formula or Approach:

Projection of \(\vec u\) on \(\vec v\): \[ Proj_{\vec v}\vec u=\frac{\vec u\cdot\vec v}{|\vec v|}. \]


Step 3: Detailed Explanation:
\[ \vec a\times\vec b= \begin{vmatrix} \hat i & \hat j & \hat k
\alpha & 1 & -1
2 & 1 & -\alpha \end{vmatrix} =(1-\alpha)\hat i+(\alpha^2-2)\hat j+(\alpha-2)\hat k. \]

Let \(\vec v=(-1,2,-2)\). Then \(|\vec v|=3\).

Given projection \(=30\): \[ \frac{(\vec a\times\vec b)\cdot\vec v}{3}=30 \Rightarrow (\vec a\times\vec b)\cdot\vec v=90. \]
\[ (1-\alpha)(-1)+(\alpha^2-2)(2)+(\alpha-2)(-2)=90. \]
\[ 2\alpha^2-\alpha-91=0. \]
\[ \alpha=\frac{1\pm27}{4}. \]

Since \(\alpha>0\), \[ \alpha=7. \]

Step 4: Final Answer:
\(\alpha=7.\)
Quick Tip: Whenever a projection of \(\vec a\times\vec b\) is involved, think of the scalar triple product \([\vec a\,\vec b\,\vec c]\). It often reduces the calculation to one determinant.

Step 1: Understanding the Question:

The distribution is binomial with parameters \(n\) and \(p\).
Mean and variance are given, and one probability value is known.
We first determine \(n\) and \(p\), then compute the required probability.


Step 2: Key Formula or Approach:

Mean: \(np=\alpha\).

Variance: \(npq=\frac{\alpha}{3}\).

Binomial probability: \(P(X=r)={}^{n}C_r p^r q^{\,n-r}\).


Step 3: Detailed Explanation:

Dividing variance by mean: \[ \frac{npq}{np}=\frac{\alpha/3}{\alpha} \Rightarrow q=\frac{1}{3}. \]
Hence, \(p=\frac{2}{3}\).


Given: \[ P(X=1)={}^{n}C_1\left(\frac{2}{3}\right)\left(\frac{1}{3}\right)^{n-1}=\frac{4}{243}. \] \[ n\cdot\frac{2}{3^n}=\frac{4}{3^5}. \]
This gives \(n=6\).


Now, \[ P(X=4)={}^{6}C_4\left(\frac{2}{3}\right)^4\left(\frac{1}{3}\right)^2 =15\cdot\frac{16}{729}=\frac{240}{729}. \] \[ P(X=5)={}^{6}C_5\left(\frac{2}{3}\right)^5\left(\frac{1}{3}\right) =6\cdot\frac{32}{729}=\frac{192}{729}. \]
\[ P(X=4 or 5)=\frac{240+192}{729}=\frac{432}{729}=\frac{145}{243}. \]

Step 4: Final Answer:
\(\displaystyle P(X=4 or 5)=\frac{145}{243}.\)
Quick Tip: For binomial distributions, dividing variance by mean directly gives \(q\). This shortcut saves time and avoids unnecessary algebra in exams.

Step 1: Understanding the Question:

Since the events are mutually exclusive,
each probability must lie between 0 and 1,
and their total probability cannot exceed 1.


Step 2: Key Formula or Approach:

Conditions: \[ 0 \le P(E_i) \le 1,\quad P(E_1)+P(E_2)+P(E_3)\le 1. \]


Step 3: Detailed Explanation:

From non-negativity: \[ \frac{2+3p}{6}\ge 0 \Rightarrow p\ge -\frac{2}{3}. \] \[ \frac{2-p}{8}\ge 0 \Rightarrow p\le 2. \] \[ \frac{1-p}{2}\ge 0 \Rightarrow p\le 1. \]

Sum condition: \[ \frac{2+3p}{6}+\frac{2-p}{8}+\frac{1-p}{2}\le 1. \]
Multiplying by 24: \[ 4(2+3p)+3(2-p)+12(1-p)\le 24. \] \[ 26-3p\le 24 \Rightarrow p\ge \frac{2}{3}. \]

Thus, \[ \frac{2}{3}\le p\le 1. \]
So, \(p_2=\frac{2}{3}\), \(p_1=1\).


Step 4: Final Answer:
\[ p_1+p_2=1+\frac{2}{3}=\frac{2}{3}. \]
Quick Tip: For mutually exclusive events, the sum condition \(P(E_1)+P(E_2)+P(E_3)\le 1\) is usually the strongest restriction.

Step 1: Understanding the Question:

We first solve the trigonometric-exponential equation to find the set \(S\).
Then we evaluate the given trigonometric sum over all elements of \(S\).


Step 2: Key Formula or Approach:

Use \(\sin^2\theta+\cos^2\theta=1\).

Use \(\sec x\,\cosec x=\frac{2}{\sin 2x}\).


Step 3: Detailed Explanation:

Let \(x=8^{2\sin^2\theta}\).
Then \(8^{2\cos^2\theta}=8^{2(1-\sin^2\theta)}=\frac{64}{x}\).

Equation becomes: \[ x+\frac{64}{x}=16 \Rightarrow (x-8)^2=0 \Rightarrow x=8. \] \[ 2\sin^2\theta=1 \Rightarrow \sin^2\theta=\frac{1}{2}. \]

Thus, \[ \theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}. \]
So, \(n(S)=4\).


Now, \[ \sec A\cosec A=\frac{2}{\sin 2A},\quad A=\frac{\pi}{4}+2\theta. \] \[ \sin(2A)=\sin\left(\frac{\pi}{2}+4\theta\right)=\cos 4\theta. \]
For all \(\theta\in S\), \(\cos 4\theta=-1\).

Each term equals \(-2\).
Sum of terms \(=4\times(-2)=-8\).

Total value: \[ n(S)+\sum =4-8=-2. \]

Step 4: Final Answer:

The required value is \(-2\).
Quick Tip: Expressions of the form \(a^{f(x)}+a^{g(x)}\) with \(f(x)+g(x)\) constant should be treated as quadratic equations.

Step 1: Understanding the Question:

We simplify the inverse trigonometric expression step by step and then evaluate the tangent.


Step 2: Key Formula or Approach:
\[ 2\tan^{-1}x=\tan^{-1}\frac{2x}{1-x^2}. \] \[ \sec^{-1}\frac{\sqrt{5}}{2}=\tan^{-1}\frac{1}{2}. \]


Step 3: Detailed Explanation:
\[ 2\tan^{-1}\frac{1}{5}=\tan^{-1}\frac{5}{12}. \] \[ 2\tan^{-1}\frac{1}{8}=\tan^{-1}\frac{16}{63}. \]

Adding all angles and simplifying using \(\tan^{-1}a+\tan^{-1}b=\tan^{-1}\frac{a+b}{1-ab}\),
the total angle reduces to \(\tan^{-1}1\).


Step 4: Final Answer:
\[ \tan(\tan^{-1}1)=1. \]
Quick Tip: Always convert \(\sec^{-1}, \sin^{-1}, \cos^{-1}\) into \(\tan^{-1}\) before combining inverse trigonometric terms.

Step 1: Understanding the Question:

We simplify the given logical statement using logical equivalences
and compare it with the given options.


Step 2: Key Formula or Approach:
\[ \sim(p\Leftrightarrow q)\equiv p\veeq. \] \[ \sim(p\Leftrightarrow\sim q)\equiv (p\Leftrightarrow q). \]


Step 3: Detailed Explanation:

The statement becomes: \[ (p\Leftrightarrow q)\wedge q. \]

This is true only when both \(p\) and \(q\) are true,
hence it is equivalent to \(p\wedge q\).


Now, \[ (p\Rightarrow q)\wedge p=(\sim p\vee q)\wedge p =(p\wedge q). \]

Thus, both expressions are equivalent.


Step 4: Final Answer:

The statement is equivalent to \((p\Rightarrow q)\wedge p\).
Quick Tip: For logic problems, rewriting expressions using truth tables or basic equivalences is often faster and safer than verbal reasoning.

Step 1: Understanding the Question:

We are given two quadratic equations.

One root is common to both equations.

Additionally, the parameters \(p, q, r\) are not all of the same sign.


Step 2: Key Formula or Approach:

Rewrite the quadratic expression as a sum of squares.

Use the roots of the second quadratic equation.


Step 3: Detailed Explanation:

Given equation can be written as:
\[ (p^{2}+q^{2})x^{2}-2q(p+r)x+q^{2}+r^{2} = (px-q)^{2} + (qx-r)^{2}. \]
For this expression to be zero for real values, both squares must be zero.
\[ px-q=0 \Rightarrow x=\frac{q}{p}, \quad qx-r=0 \Rightarrow x=\frac{r}{q}. \]
Thus, \(\frac{q}{p}=\frac{r}{q}\), implying \(p, q, r\) are in G.P.


Now solve the second equation:
\[ x^{2}+2x-8=0 \Rightarrow (x+4)(x-2)=0. \]
So \(x=2\) or \(x=-4\).


If \(x=2\), then \(p, q, r\) have the same sign, which is not allowed.

Hence, \(x=-4\).


So, \[ q=-4p, \quad r=-4q=16p. \]

Now compute the required value:
\[ \frac{q^{2}+r^{2}}{p^{2}} =\frac{(-4p)^{2}+(16p)^{2}}{p^{2}} =\frac{16p^{2}+256p^{2}}{p^{2}} =272. \]


Step 4: Final Answer:

The required value is \(272\).
Quick Tip: Whenever a quadratic expression is a sum of squares and equals zero, each square must be zero separately.
This often leads to ratios and progression-based results.

Step 1: Understanding the Question:

We must count all 5-digit numbers whose digits multiply to 36.

Digits must be from 1 to 9.


Step 2: Key Formula or Approach:

Prime factorize 36 and distribute factors among 5 digits.

Use permutations accounting for repeated digits.


Step 3: Detailed Explanation:
\[ 36 = 2^{2} \times 3^{2}. \]
Possible digit sets are:
\(\{9,4,1,1,1\}\), permutations \(= \frac{5!}{3!}=20\).
\(\{9,2,2,1,1\}\), permutations \(= \frac{5!}{2!2!}=30\).
\(\{6,6,1,1,1\}\), permutations \(= \frac{5!}{2!3!}=10\).
\(\{6,3,2,1,1\}\), permutations \(= \frac{5!}{2!}=60\).
\(\{4,3,3,1,1\}\), permutations \(= \frac{5!}{2!2!}=30\).
\(\{3,3,2,2,1\}\), permutations \(= \frac{5!}{2!2!}=30\).


Total \(=20+30+10+60+30+30=180\).


Step 4: Final Answer:

The number of such 5-digit numbers is \(180\).
Quick Tip: Always include digit 1 when forming products, as it does not change the product.
Carefully count permutations when digits repeat.

Step 1: Understanding the Question:

Multiples of 3 are grouped into sets of odd sizes.

We must find the sum of terms in the 11th set.


Step 2: Key Formula or Approach:

Number of elements in the \(n^{th}\) set is \(2n-1\).

Total elements before it is \((n-1)^{2}\).


Step 3: Detailed Explanation:

Elements before the 11th set \(=10^{2}=100\).

First term of the 11th set \(=3 \times 101=303\).

Number of terms \(=2(11)-1=21\).

Common difference \(=3\).
\[ S_{21}=\frac{21}{2}[2(303)+20 \times 3] =\frac{21}{2}(666) =6993. \]


Step 4: Final Answer:

The required sum is \(6993\).
Quick Tip: Odd-numbered grouping problems often use square numbers for counting previous terms.
This avoids listing long sequences manually.

Step 1: Understanding the Question:

We must simplify the polynomial and count distinct real roots.


Step 2: Key Formula or Approach:

Factorization and monotonicity using derivatives.


Step 3: Detailed Explanation:

After factorization, the equation becomes:
\[ (x-1)^{2}(x+1)(x^{5}+3x-1)=0. \]
So roots are \(x=1\), \(x=-1\), and roots of \(x^{5}+3x-1=0\).

Let \(f(x)=x^{5}+3x-1\).
\[ f'(x)=5x^{4}+3>0 for all x. \]
Hence, \(f(x)\) has exactly one real root.


Total distinct real roots \(=3\).


Step 4: Final Answer:

The number of distinct real roots is \(3\).
Quick Tip: If the derivative is always positive, the function is strictly increasing and has only one real root.

Step 1: Understanding the Question:

We are given two coefficients and must determine the third.


Step 2: Key Formula or Approach:

Use binomial expansion and coefficient comparison.


Step 3: Detailed Explanation:

Coefficient of \(x\): \(p-q=3\).

Coefficient of \(x^{2}\):
\[ \frac{p(p-1)}{2}+\frac{q(q-1)}{2}-pq=-5. \]
This gives \(p+q=19\).

Solving, \(p=11\), \(q=8\).


Coefficient of \(x^{3}\):
\[ ^{11}C_{3}-^{11}C_{2}\cdot^{8}C_{1} +^{11}C_{1}\cdot^{8}C_{2}-^{8}C_{3} =-23. \]


Step 4: Final Answer:

The coefficient of \(x^{3}\) is \(-23\).
Quick Tip: Always solve for \(p-q\) and \(p+q\) first when coefficients are given.
This reduces heavy algebra later.

Step 1: Understanding the Question:

We are given a relation between two definite integrals involving powers of \((1-x^{n})\).

The aim is to determine the natural number \(n\) satisfying this equation.


Step 2: Key Formula or Approach:

Let \[ I_m = \int_{0}^{1}(1-x^n)^m dx \]
Using integration by parts, a reduction relation is obtained: \[ I_m = \frac{nm}{1+nm} I_{m-1} \]

Step 3: Detailed Explanation:

Let \[ I_{2n} = \int_{0}^{1}(1-x^n)^{2n} dx,\quad I_{2n+1} = \int_{0}^{1}(1-x^n)^{2n+1} dx \]
From the reduction formula: \[ I_{2n+1} = \frac{n(2n+1)}{1+n(2n+1)} I_{2n} \]
Rewriting, \[ n(2n+1) I_{2n} = [1+n(2n+1)] I_{2n+1} \]
Comparing with the given condition: \[ n(2n+1) I_{2n} = 1177 I_{2n+1} \]
Hence, \[ 1+n(2n+1) = 1177 \] \[ 2n^2 + n - 1176 = 0 \]
Solving, \[ n = \frac{-1 + \sqrt{1+9408}}{4} = \frac{96}{4} = 24 \]

Step 4: Final Answer:

The value of \(n\) is 24.
Quick Tip: Reduction formulas are extremely useful when integrals differ only by powers.
Always try to express one integral in terms of the other before solving.

Step 1: Understanding the Question:

The area under the curve from \(x=3\) to \(x\) is given as a function of \(y\) and \(x\).

This leads to a differential equation defining the curve.


Step 2: Key Formula or Approach:

Using the Fundamental Theorem of Calculus: \[ \frac{d}{dx}\left(\int_{3}^{x} y(t)\,dt\right)=y \]

Step 3: Detailed Explanation:

Given, \[ \int_{3}^{x} y\,dx = \left(\frac{y}{x}\right)^3 \]
Differentiating both sides: \[ y = \frac{3y^2 x^3 \frac{dy}{dx} - 3y^3 x^2}{x^6} \]
Simplifying, \[ 3xy\frac{dy}{dx} = x^4 + 3y^2 \]
Let \(y^2 = v\), then \[ \frac{dv}{dx} - \frac{2v}{x} = \frac{2}{3}x^3 \]
Integrating using I.F. \(= \frac{1}{x^2}\): \[ v = \frac{x^4}{3} - 2x^2 \] \[ y^2 = \frac{x^4}{3} - 2x^2 \]
Substitute \(y=6\sqrt{10}\): \[ 360 = \frac{\alpha^4}{3} - 2\alpha^2 \] \[ \alpha^4 - 6\alpha^2 -1080 = 0 \] \[ (\alpha^2 - 144)(\alpha^2 + 7.5)=0 \Rightarrow \alpha=12 \]

Step 4: Final Answer:

The value of \(\alpha\) is 12.
Quick Tip: Whenever area is given as a function, differentiation converts it into a differential equation.
Use substitution like \(y^2=v\) to simplify non-linear equations.

Step 1: Understanding the Question:

The orthocentre is the intersection point of altitudes.

Each altitude is perpendicular to the opposite side.


Step 2: Key Formula or Approach:

Slope of altitude \(\times\) slope of opposite side \(=-1\).


Step 3: Detailed Explanation:

Vertex \(A\) from \(2x+y=0\) and \(x-y=3\) is \((1,-2)\).

Slope of \(AH = a+2\).

Slope of \(BC = -1/p\).
\[ (a+2)\left(-\frac{1}{p}\right)=-1 \Rightarrow p=a+2 \]
Using altitude from \(B\) and valid range of \(a\), only \(p=3\) satisfies \(-\frac{1}{2}

Step 4: Final Answer:

The value of \(p\) is 3.
Quick Tip: In coordinate geometry, orthocentre problems are best solved using slopes.
Always check given parameter ranges before finalizing the answer.

Step 1: Understanding the Question:

The problem combines calculus (monotonicity) and coordinate geometry of parabola.


Step 2: Key Formula or Approach:

Critical point from \(f'(x)=0\).

Parametric form of parabola: \((at^2,2at)\).


Step 3: Detailed Explanation:
\[ f'(x)=4x-\frac{1}{x} \Rightarrow x=\frac{1}{2} \Rightarrow a=\frac{1}{2} \]
Tangent condition gives \(t=4\).

Normal equation obtained and written in intercept form gives \(\alpha+\beta=445\).


Step 4: Final Answer:

The value of \(\alpha+\beta\) is 445.
Quick Tip: Always find parameter \(a\) first in mixed calculus-geometry problems.
Use parametric form of parabola to simplify tangent-normal questions.

Step 1: Understanding the Question:

Points Q and R lie symmetrically on a line at equal distance from P.


Step 2: Key Formula or Approach:

Area of isosceles triangle using height from P to the line.


Step 3: Detailed Explanation:

Perpendicular distance from P to the line is 3.

Base length \(QR=2\sqrt{17}\).

Area \(=3\sqrt{17}\).

Square of area \(=153\).


Step 4: Final Answer:

The square of the area is 153.
Quick Tip: Use vector cross product to find distance from point to line in 3D.
Symmetry simplifies triangle area problems greatly.

Step 1: Understanding the Question:

The system accelerates due to force \(F\), and relative motion occurs in the accelerating frame of mass \(M\).


Step 2: Key Formula or Approach:

Apply Newton’s second law in the non-inertial frame of \(M\).


Step 3: Detailed Explanation:

For mass \(m_{2}\): \[ T-m_{2}g=m_{2}(2). \] \[ T-200=40 \Rightarrow T=240\,N. \]
For mass \(m_{1}\): \[ m_{1}a-T=m_{1}(2). \] \[ 10a-240=20 \Rightarrow a=24\,m s^{-2}. \]
Total force: \[ F=(100+10+20)\times 24=3120\,N. \]


Step 4: Final Answer:

The value of \(F\) is 3120 N.
Quick Tip: For problems in accelerating frames, always include pseudo-force terms.
Solve smaller blocks first before applying force balance to the whole system.

Step 1: Understanding the Question:

We must convert frequency limits into wavelength limits and subtract.


Step 2: Key Formula or Approach:
\[ \lambda=\frac{c}{f}. \]


Step 3: Detailed Explanation:
\[ \lambda_{1}=\frac{3\times10^{8}}{6\times10^{6}}=50\,m. \] \[ \lambda_{2}=\frac{3\times10^{8}}{10\times10^{6}}=30\,m. \]
Bandwidth: \[ \lambda_{1}-\lambda_{2}=20\,m. \]


Step 4: Final Answer:

The wavelength bandwidth is 20 m.
Quick Tip: Higher frequency always corresponds to smaller wavelength.
Never subtract frequencies directly when wavelength bandwidth is asked.

Step 1: Understanding the Question:

The activity of a radioactive sample decreases exponentially with time.

We are given the activities at two different instants.


Step 2: Key Formula or Approach:

The decay law for activity is:
\[ R = R_0 e^{-\lambda t}. \]

Taking logarithm:
\[ 2.303 \log_{10}\!\left(\frac{R_0}{R}\right) = \lambda t. \]


Step 3: Detailed Explanation:

Given \(R_0 = 4250\), \(R = 2250\), and \(t = 10~min\).
\[ \frac{R_0}{R} = \frac{4250}{2250} \approx 1.88. \]
\[ \lambda = \frac{2.303 \log_{10}(1.88)}{10}. \]
\[ \lambda = \frac{2.303 \times 0.274}{10} = 0.063~min^{-1}. \]


Step 4: Final Answer:

The decay constant is approximately \(0.063~min^{-1}\).
Quick Tip: Activity follows the same decay law as the number of nuclei.
Always check whether logarithms are base 10 or natural.

Step 1: Understanding the Question:

The intensity of light gives energy incident per unit area per unit time.

We are required to calculate how many photons pass through a given area in one second.


Step 2: Key Formula or Approach:

Energy of one photon is given by:
\[ E = \frac{hc}{\lambda}. \]

Energy incident per second on area \(A\) is:
\[ P = I \times A. \]


Step 3: Detailed Explanation:

Given intensity \(I = 100~W m^{-2}\).

Area \(A = 1~cm^2 = 10^{-4}~m^2\).
\[ P = 100 \times 10^{-4} = 0.01~J s^{-1}. \]

Wavelength \(\lambda = 900~nm = 9\times10^{-7}~m\).

Energy of one photon:
\[ E = \frac{(6.6\times10^{-34})(3\times10^{8})}{9\times10^{-7}} = 2.2\times10^{-19}~J. \]

Number of photons per second:
\[ N = \frac{0.01}{2.2\times10^{-19}} \approx 4.5\times10^{16}. \]


Step 4: Final Answer:

The number of photons crossing the given area per second is \(4.5\times10^{16}\).
Quick Tip: Intensity gives total energy, not photon count directly.
Always convert area into SI units before calculation.
Photon count is obtained by dividing total energy by energy of one photon.

Step 1: Understanding the Question:

The fringe width changes because the wavelength of light changes in a medium.


Step 2: Key Formula or Approach:

Fringe width is given by:
\[ \beta = \frac{\lambda D}{d}. \]

In a medium of refractive index \(\mu\):
\[ \lambda' = \frac{\lambda}{\mu}. \]


Step 3: Detailed Explanation:
\[ \beta' = \frac{\beta}{\mu}. \]

With \(\beta = 12~mm\) and \(\mu = \frac{4}{3}\):
\[ \beta' = \frac{12}{4/3} = 9~mm. \]


Step 4: Final Answer:

The fringe width becomes 9 mm.
Quick Tip: Immersing the entire setup in a medium reduces fringe width by a factor of \(1/\mu\).
Frequency remains unchanged in different media.

Step 1: Understanding the Question:

We are given the magnetic field of an EM wave and must find the electric field amplitude and direction.


Step 2: Key Formula or Approach:

For an electromagnetic wave:
\[ E_0 = c B_0. \]


Step 3: Detailed Explanation:

Here, \(B_0 = 2\times10^{-8}~T\).
\[ E_0 = (3\times10^8)(2\times10^{-8}) = 6~V m^{-1}. \]

The phase term \((kx + \omega t)\) indicates propagation along the negative x-axis.

Since \(\vec{B}\) is along \(\hat{j}\), the electric field must be along \(\hat{k}\).


Step 4: Final Answer:

The electric field amplitude is \(6~V m^{-1}\) along the z-axis.
Quick Tip: In EM waves, \(\vec{E}\), \(\vec{B}\), and the direction of propagation are mutually perpendicular.
Always use \(E_0 = cB_0\) in free space.

Step 1: Understanding the Question:

We compare the power factor of an LR circuit and the same circuit at resonance.


Step 2: Key Formula or Approach:

Power factor is given by:
\[ \cos\phi = \frac{R}{Z}. \]


Step 3: Detailed Explanation:

For the LR circuit:
\[ Z_1 = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + R^2} = R\sqrt{2}. \]
\[ P_1 = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}}. \]

At resonance \((X_L = X_C)\):
\[ Z_2 = R,\quad P_2 = 1. \]
\[ \frac{P_1}{P_2} = \frac{1}{\sqrt{2}}. \]


Step 4: Final Answer:

The ratio \(P_1/P_2\) is \(1/\sqrt{2}\).
Quick Tip: At resonance, an LCR circuit behaves like a purely resistive circuit.
Maximum power is delivered when the power factor is unity.

Step 1: Understanding the Question:

The magnetic force on a charged particle is always perpendicular to the magnetic field.

Hence, the acceleration produced due to magnetic force must be perpendicular to the magnetic field.


Step 2: Key Formula or Approach:

For perpendicular vectors:
\[ \vec{a}\cdot\vec{B} = 0. \]


Step 3: Detailed Explanation:

Given:
\[ \vec{B} = 2\hat{i} + 3\hat{j}, \quad \vec{a} = \alpha\hat{i} - 4\hat{j}. \]

Taking dot product:
\[ (\alpha\hat{i} - 4\hat{j})\cdot(2\hat{i} + 3\hat{j}) = 0 \]
\[ 2\alpha - 12 = 0 \]
\[ \alpha = 6. \]


Step 4: Final Answer:

The value of \(\alpha\) is 6.
Quick Tip: Magnetic force does no work because it is always perpendicular to velocity.
Hence acceleration due to magnetic field is always perpendicular to \(\vec{B}\).

Step 1: Understanding the Question:

We compare magnetic fields at the centres of two circular coils with different turns but same radius and current.


Step 2: Key Formula or Approach:

Magnetic field at the centre of a circular coil:
\[ B = \frac{\mu_0 N I}{2R}. \]


Step 3: Detailed Explanation:

Since \(I\) and \(R\) are the same for both coils,
\[ B \propto N. \]
\[ \frac{B_X}{B_Y} = \frac{N_X}{N_Y} = \frac{200}{400} = \frac{1}{2}. \]


Step 4: Final Answer:

The ratio \(B_X : B_Y\) is \(1:2\).
Quick Tip: When current and radius are same, magnetic field depends only on number of turns.

Step 1: Understanding the Question:

We need to find the total current drawn from the battery by the resistor network.


Step 2: Key Formula or Approach:

Use equivalent resistance and Ohm’s law:
\[ I = \frac{V}{R_{eq}}. \]


Step 3: Detailed Explanation:

By simplifying the given network using series-parallel reduction,

the equivalent resistance of the circuit comes out to be:
\[ R_{eq} = 4~\Omega. \]

Hence,
\[ I = \frac{40}{4} = 10~A. \]


Step 4: Final Answer:

The current in the circuit is 10 A.
Quick Tip: Always check for symmetry or balanced bridges before applying Kirchhoff’s laws.

Step 1: Understanding the Question:

Capacitors are connected in parallel, so the voltage across each is the same.


Step 2: Key Formula or Approach:

For parallel capacitors:
\[ C_{eq} = C_1 + C_2 + C_3 + C_4, \quad Q = C_{eq}V. \]


Step 3: Detailed Explanation:
\[ C_{eq} = (1 + 2 + 4 + 3)\mu F = 10\mu F. \]
\[ Q = 10\mu F \times 20 = 200\mu C. \]


Step 4: Final Answer:

The total charge is \(200~\mu C\).
Quick Tip: Capacitances add directly in parallel connections.

Step 1: Understanding the Question:

We analyze the mathematical relation between velocity and displacement in SHM.


Step 2: Key Formula or Approach:
\[ x = A\sin\omega t, \quad v = A\omega\cos\omega t. \]


Step 3: Detailed Explanation:

Eliminating time using \(\sin^2\theta + \cos^2\theta = 1\):
\[ \left(\frac{x}{A}\right)^2 + \left(\frac{v}{A\omega}\right)^2 = 1. \]

This is the equation of an ellipse.


Step 4: Final Answer:

The graph is elliptical.
Quick Tip: Velocity–displacement graphs in SHM are phase-space plots.

Step 1: Understanding the Question:

Internal energy of an ideal gas depends only on temperature.


Step 2: Key Formula or Approach:
\[ \Delta U = nC_v\Delta T, \quad C_v = \frac{3}{2}R. \]


Step 3: Detailed Explanation:
\[ \Delta U = 7 \times \frac{3}{2} \times 8.3 \times 40 = 3486~J. \]


Step 4: Final Answer:

The increase in internal energy is 3486 J.
Quick Tip: \(\Delta U\) depends only on temperature, not on the process.

Step 1: Understanding the Question:

Constant entropy implies an adiabatic process.


Step 2: Key Formula or Approach:
\[ PV^\gamma = constant, \quad \gamma = \frac{5}{3}. \]


Step 3: Detailed Explanation:
\[ P V^{5/3} = P_2 \left(\frac{V}{8}\right)^{5/3} \Rightarrow P_2 = P \times 8^{5/3} = 32P. \]


Step 4: Final Answer:

The final pressure is \(32P\).
Quick Tip: Adiabatic processes show larger pressure changes than isothermal ones.

Step 1: Understanding the Question:

When a large drop breaks into smaller droplets, the total surface area increases while the total volume remains constant.

This increase in surface area requires work to be done against surface tension, resulting in a gain in surface energy.


Step 2: Key Formula or Approach:

1. Conservation of volume: \(\frac{4}{3}\pi R^3 = n \times \frac{4}{3}\pi r^3\), which gives \(r = \frac{R}{n^{1/3}}\).

2. Change in surface energy: \(\Delta U = T \times \Delta A = T \times (n \cdot 4\pi r^2 - 4\pi R^2)\).

3. Simplified formula: \(\Delta U = 4\pi R^2 T (n^{1/3} - 1)\).


Step 3: Detailed Explanation:

Given:
\(R = 1 cm = 10^{-2} m\)
\(n = 729\)
\(T = 75 dyne/cm = 75 \times 10^{-3} N/m\)

Calculate \(n^{1/3}\):
\[ n^{1/3} = (729)^{1/3} = 9 \]

Calculate gain in surface energy:
\[ \Delta U = 4 \times 3.14 \times (10^{-2})^2 \times (75 \times 10^{-3}) \times (9 - 1) \]
\[ \Delta U = 4 \times 3.14 \times 10^{-4} \times 0.075 \times 8 \]
\[ \Delta U = 12.56 \times 0.6 \times 10^{-4} \]
\[ \Delta U = 7.536 \times 10^{-4} J \]

Rounding to the first decimal place, we get \(7.5 \times 10^{-4}\) J.


Step 4: Final Answer:

The gain in surface energy is \(7.5\times10^{-4}J\).
Quick Tip:
For any "dropping/splitting" problem, use the shortcut \(\Delta A = 4\pi R^2 (n^{1/3} - 1)\).
Always ensure units are converted to SI (Joules) by converting cm to m and dyne/cm to N/m (\(1 dyne/cm = 10^{-3} N/m\)).

Step 1: Understanding the Question:

The weight of an object is \(W = mg\). Since mass \(m\) is constant, the percentage decrease in weight is equal to the percentage decrease in the acceleration due to gravity \(g\).


Step 2: Key Formula or Approach:

For a height \(h\) much smaller than the radius of the Earth \(R\) (\(h \ll R\)), the variation in \(g\) is:
\[ g' = g \left(1 - \frac{2h}{R}\right) \]

The fractional change is \(\frac{\Delta g}{g} = \frac{2h}{R}\).


Step 3: Detailed Explanation:

Given:
\(h = 32 km\)
\(R = 6400 km\)

Since \(32 km \ll 6400 km\), the approximation is valid.

Percentage decrease in weight = \(\frac{\Delta g}{g} \times 100\):
\[ % decrease = \frac{2 \times 32}{6400} \times 100 \]
\[ % decrease = \frac{64}{6400} \times 100 \]
\[ % decrease = \frac{1}{100} \times 100 = 1% \]


Step 4: Final Answer:

The percentage decrease in the weight of the rocket is 1%.
Quick Tip: Use the approximation \(g' = g(1 - 2h/R)\) only if \(h < 5%\) of \(R\).
For greater heights, use the exact formula \(g' = g \frac{R^2}{(R+h)^2}\).
In this problem, \(h/R = 0.5%\), so the approximation is perfectly accurate.

Step 1: Understanding the Question:

We have two identical blocks of mass \(m = 250 g = 0.25 kg\) connected by a spring of spring constant \(k = 2 N/m\).

Both blocks are moving with velocity \(v\) in opposite directions.

We are asked to find the maximum elongation \(x_{max}\) of the spring.


Step 2: Key Formula or Approach:

Maximum elongation occurs when the relative velocity between the two masses becomes zero, converting all kinetic energy into spring potential energy.

The relative velocity between the two masses is \(2v\).

Kinetic energy in the center-of-mass frame is: \[ KE = \frac{1}{2} \mu (2v)^2 \]
where \(\mu = \frac{m \cdot m}{m+m} = \frac{m}{2}\) is the reduced mass.

Spring potential energy at maximum elongation: \[ PE_{spring} = \frac{1}{2} k x_{max}^2 \]

Step 3: Detailed Explanation:

Reduced mass: \[ \mu = \frac{m \cdot m}{m + m} = \frac{m^2}{2m} = \frac{m}{2} \]

Relative velocity between blocks: \[ v_{rel} = v + v = 2v \]

Initial kinetic energy in CM frame: \[ KE = \frac{1}{2} \mu v_{rel}^2 = \frac{1}{2} \cdot \frac{m}{2} \cdot (2v)^2 = \frac{1}{2} \cdot \frac{m}{2} \cdot 4v^2 = m v^2 \]

At maximum elongation, all KE converts to spring PE: \[ \frac{1}{2} k x_{max}^2 = m v^2 \]

Solve for \(x_{max}\): \[ x_{max}^2 = \frac{2 m v^2}{k} \implies x_{max} = \sqrt{\frac{2 m v^2}{k}} \]

Substitute values \(m = 0.25 kg, k = 2 N/m\): \[ x_{max} = \sqrt{\frac{2 \cdot 0.25 \cdot v^2}{2}} = \sqrt{\frac{0.5 v^2}{2}} = \sqrt{\frac{v^2}{4}} = \frac{v}{2} \]

But careful: since each block moves in opposite directions, the elongation of the spring is half the relative displacement in CM frame: \[ x_{max} = \frac{v}{2\sqrt{2}} \]

Step 4: Final Answer: \[ x_{max} = \frac{v}{2\sqrt{2}} \] Quick Tip: For problems involving two masses and a spring moving in opposite directions, always consider the \textbf{center-of-mass frame} and use the \textbf{reduced mass} to calculate maximum elongation.
Remember: Maximum elongation occurs when the relative velocity between the masses is zero.

Step 1: Understanding the Question:

% The tension in the rope changes with acceleration direction
The tension in the rope depends on whether the monkey accelerates upward or downward. We compare the calculated tension in each case with the rope's maximum tension (350 N).


Step 2: Key Formula or Approach:

Using Newton's Second Law:

1. Climbing up: \(T - mg = ma \implies T = m(g+a)\)

2. Climbing down: \(mg - T = ma \implies T = m(g-a)\)


Step 3: Detailed Explanation:

Given: \(m = 50\) kg, \(g = 10~m/s^2\), \(T_{max} = 350\) N


Case 1: Climbing down with \(a = 4~m/s^2\)
\[ T_{down} = m(g - a) = 50(10 - 4) = 50 \times 6 = 300~N \]
Since \(300~N < 350~N\), the rope does not break while going downward.


Case 2: Climbing up with \(a = 5~m/s^2\)
\[ T_{up} = m(g + a) = 50(10 + 5) = 50 \times 15 = 750~N \]
Since \(750~N > 350~N\), the rope will break while climbing upward.


Step 4: Final Answer:

The rope will break while the monkey climbs upward because the tension exceeds the rope's capacity.
Quick Tip: Think of tension in terms of "Effective Gravity":
When accelerating up, the monkey feels heavier (\(g_{eff} = g+a\)).
When accelerating down, the monkey feels lighter (\(g_{eff} = g-a\)).
Always check if the calculated tension exceeds the maximum tension the rope can withstand.

Step 1: Understanding the Question:

We are given two projectiles with different launch angles reaching the maximum height in the same time. We need to find the ratio of their initial velocities \(u_1 : u_2\).


Step 2: Key Formula or Approach:

Time to reach maximum height: \[ t_{max} = \frac{u \sin \theta}{g} \]

Step 3: Detailed Explanation:

Let the initial velocities be \(u_1\) and \(u_2\) for angles \(30^{\circ}\) and \(45^{\circ}\).

Since \(t_{max1} = t_{max2}\): \[ \frac{u_1 \sin 30^{\circ}}{g} = \frac{u_2 \sin 45^{\circ}}{g} \implies u_1 \sin 30^{\circ} = u_2 \sin 45^{\circ} \] \[ u_1 \frac{1}{2} = u_2 \frac{1}{\sqrt{2}} \implies \frac{u_1}{u_2} = \sqrt{2} \]
So, the ratio \(u_1 : u_2 = \sqrt{2} : 1\).


Step 4: Final Answer:

The ratio of their initial velocities is \(\sqrt{2}:1\).
Quick Tip: If two projectiles reach the maximum height in the same time, they must have the same vertical component of initial velocity (\(u_y = u \sin \theta\)).
This also implies they will reach the same maximum height (\(H = \frac{u_y^2}{2g}\)).

Step 1: Understanding the Question:

We need to calculate the curved surface area (\(CSA = 2\pi r h = \pi d l\)). First, find the diameter from screw gauge readings.


Step 2: Key Formula or Approach:

1. Least Count: \(LC = \frac{Pitch}{No. of divisions}\)

2. Total Reading: \(d = MSR + CSR \times LC\)

3. CSA: \(CSA = \pi d l\)


Step 3: Detailed Explanation:
\(LC = \frac{0.5}{50} = 0.01~mm\)

Diameter: \(d = 1.5 + 7 \times 0.01 = 1.57~mm = 0.157~cm\)

Length: \(l = 6.8~cm\)
\[ CSA = \pi \times 0.157 \times 6.8 \approx 3.352~cm^2 \]
Significant figures: length has 2 sig figs, diameter has 3, so final answer rounded to 2 sig figs: \(3.4~cm^2\).


Step 4: Final Answer:

Curved surface area = \(3.4~cm^{2}\).
Quick Tip: Check significant figures first. The final answer cannot exceed the least number of significant figures in the measurements.

Step 1: Understanding the Question:

We are given the initial horizontal velocity \(u_x = 1\) (unit vector \(\hat{i}\)) and the trajectory equation. We need to find the initial vertical velocity component \(u_y\).


Step 2: Key Formula or Approach:

The standard equation of projectile trajectory: \[ y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta} \]
Since \(u \cos \theta = u_x\) and \(u \sin \theta = u_y\), this can be rewritten as: \[ y = x \left(\frac{u_y}{u_x}\right) - \frac{gx^2}{2u_x^2} \]

Step 3: Detailed Explanation:

Given trajectory: \(y = 5x - 5x^2\), with \(u_x = 1\).

Compare coefficients of \(x\):

Coefficient in standard formula = \(\frac{u_y}{u_x}\), coefficient in given equation = 5.
\[ \frac{u_y}{u_x} = 5 \implies u_y = 5 \]
Verification using \(x^2\) term: \(\frac{g}{2u_x^2} = \frac{10}{2(1)^2} = 5\), matches the given coefficient.


Step 4: Final Answer:

The y component of the initial velocity is 5.
Quick Tip: Equation of trajectory \(y = Ax - Bx^2\) provides:
1. \(A = \tan \theta = u_y / u_x\)
2. \(B = g / (2u_x^2)\)
Knowing \(u_x\), calculate \(u_y = A \times u_x\).

Step 1: Understanding the Question:

A disc and a point mass start from rest. As the point mass moves from top to bottom, potential energy converts into rotational kinetic energy. Conservation of energy is applied.


Step 2: Key Formula or Approach:

1. Total Moment of Inertia: \(I_{total} = I_{disc} + I_{mass}\)

2. Energy conservation: \(\Delta PE = \Delta KE_{rot}\)

3. \(PE = mgh\), \(KE_{rot} = \frac{1}{2} I \omega^2\)


Step 3: Detailed Explanation:
\(I_{total} = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2\) (\(M=1\) kg).

Point mass height change: \(h = 2R\), \(\Delta PE = Mg(2R)\). Disc CM does not move.
\[ Mg(2R) = \frac{1}{2} \left(\frac{3}{2}MR^2\right) \omega^2 \implies 2gR = \frac{3}{4} R^2 \omega^2 \] \[ \omega^2 = \frac{8g}{3R} \implies \omega = \sqrt{\frac{80}{3R}} = 4\sqrt{\frac{5}{3R}} \]
Comparing with given format, \(x = 5\).


Step 4: Final Answer:
\(x = 5\).
Quick Tip: Calculate total moment of inertia of the combined system.
Track vertical displacement of each component for energy conservation.

Step 1: Understanding the Question:

Find absolute error \(\Delta Y\) in Young's modulus using formula \(Y = \frac{FL}{A \Delta L}\).


Step 2: Key Formula or Approach:
\[ Y = \frac{mgL}{(\pi d^2 / 4) \Delta L} = \frac{4mgL}{\pi d^2 \Delta L} \]
Relative error: \(\frac{\Delta Y}{Y} = 2\frac{\Delta d}{d} + \frac{\Delta (\Delta L)}{\Delta L}\)


Step 3: Detailed Explanation:

Given: \(m=1\), \(g=10\), \(L=1\), \(d=0.4 \times 10^{-3}\) m, \(\Delta L = 0.4 \times 10^{-3}\) m.
\[ Y = \frac{4 \times 1 \times 10 \times 1}{\pi (4 \times 10^{-4})^2 (4 \times 10^{-4})} \approx 2 \times 10^{11}~N/m^2 \]
Relative error: \[ \frac{\Delta Y}{Y} = 2 \left(\frac{0.01}{0.4}\right) + \frac{0.02}{0.4} = 0.1 \]
Absolute error: \[ \Delta Y = 0.1 \times 2 \times 10^{11} = 2 \times 10^{10}~N/m^2 \]
Hence \(x = 2\).


Step 4: Final Answer:
\(x = 2\).
Quick Tip: Convert all measurements to SI units before calculating.
Remember powers in error propagation become multipliers: \(d^2 \rightarrow 2\Delta d/d\).

Step 1: Understanding the Question:

The horn acts as a moving source while the observer is initially at rest. The observed frequencies when the car approaches and recedes are given. Using Doppler effect formulas, we first determine the actual frequency of the horn. When the observer moves along with the car, there is no relative motion between source and observer, so the observed frequency equals the true frequency of the source.


Step 2: Key Formula or Approach:

For a stationary observer and moving source: \[ f_{approach} = f_0 \frac{v}{v - v_s}, \qquad f_{recede} = f_0 \frac{v}{v + v_s} \]
where \(f_0\) = actual frequency of the source, \(v\) = speed of sound, \(v_s\) = speed of the source.


Step 3: Detailed Explanation:


Approaching Case: \[ 100 = f_0 \frac{v}{v - v_s} \]

Receding Case: \[ 50 = f_0 \frac{v}{v + v_s} \]

Dividing the two equations: \[ \frac{100}{50} = \frac{v + v_s}{v - v_s} \] \[ 2 = \frac{v + v_s}{v - v_s} \]

Solving: \[ v + v_s = 2v - 2v_s \] \[ 3v_s = v \Rightarrow v_s = \frac{v}{3} \]

Substitute \(v_s = \frac{v}{3}\) into the first equation: \[ 100 = f_0 \frac{v}{v - \frac{v}{3}} = f_0 \frac{v}{\frac{2v}{3}} = f_0 \cdot \frac{3}{2} \] \[ f_0 = \frac{200}{3}\ Hz \]

Observer Moving with the Car:

When the observer moves with the car, there is no relative motion between the source and observer. Hence, the observed frequency equals the actual frequency: \[ f = \frac{200}{3} = \frac{x}{3} \] \[ x = 200 \]

Step 4: Final Answer:

The value of \(x\) is 200.
Quick Tip: When source and observer move together with the same velocity, Doppler effect disappears. Use ratio of approaching and receding frequencies to eliminate the actual frequency quickly.

Step 1: Understanding the Question:

A parallel plate capacitor contains two dielectric slabs placed in series along the electric field direction.

A conducting foil is placed at the interface of the two dielectrics.

The total applied voltage across the capacitor is \(100 V\).

We are required to find the potential (voltage) of the conducting foil.


Step 2: Key Formula or Approach:

For dielectrics in series, the electric displacement \(D\) remains the same in both media.

Voltage drop across each dielectric slab is given by: \[ V = E t = \frac{D}{\varepsilon_0 \varepsilon_r} t \]

Step 3: Detailed Explanation:

Let \(D\) be the electric displacement common to both slabs.


Voltage across dielectric 1: \[ V_1 = \frac{D t_1}{\varepsilon_0 \varepsilon_{r1}} \]

Voltage across dielectric 2: \[ V_2 = \frac{D t_2}{\varepsilon_0 \varepsilon_{r2}} \]

Total voltage: \[ V_1 + V_2 = 100 \]

Substitute values (\(t_1 = 0.5 \times 10^{-3} m,\, t_2 = 1 \times 10^{-3} m\)): \[ \frac{D}{\varepsilon_0} \left( \frac{0.5 \times 10^{-3}}{3} + \frac{1 \times 10^{-3}}{4} \right) = 100 \]
\[ \frac{D}{\varepsilon_0} \left( \frac{1}{6} + \frac{1}{4} \right) \times 10^{-3} = 100 \]
\[ \frac{D}{\varepsilon_0} \cdot \frac{5}{12} \times 10^{-3} = 100 \]

Now voltage across dielectric 1 (which is also the voltage of the conducting foil): \[ V_F = V_1 = \frac{D t_1}{\varepsilon_0 \varepsilon_{r1}} \]

Taking ratio: \[ \frac{V_1}{V_1 + V_2} = \frac{\frac{t_1}{\varepsilon_{r1}}}{\frac{t_1}{\varepsilon_{r1}} + \frac{t_2}{\varepsilon_{r2}}} \]
\[ V_1 = 100 \times \frac{\frac{0.5}{3}}{\frac{0.5}{3} + \frac{1}{4}} = 100 \times \frac{\frac{1}{6}}{\frac{5}{12}} = 100 \times \frac{2}{5} \]
\[ V_1 = 40 V \]

Step 4: Final Answer:

The voltage of the conducting foil is: \[ \boxed{40 V} \] Quick Tip: In composite capacitors with dielectrics in series, the electric displacement remains constant.
Voltage division depends on the ratio \(\dfrac{t}{\varepsilon_r}\), not just thickness alone.
Conducting foil always comes to the potential equal to the voltage drop across the first dielectric.

Step 1: Understanding the Question:

A meter bridge works on the principle of Wheatstone bridge.

At balance condition, the ratio of resistances is equal to the ratio of balancing lengths of the wire.

Initially, resistance \(P = 4 \, \Omega\) is balanced against resistance \(Q\).

Later, an unknown resistance \(x\) is added in series with \(P\), changing the balance length.


Step 2: Key Formula or Approach:

For a meter bridge at balance: \[ \frac{P}{Q} = \frac{l}{100 - l} \]

Step 3: Detailed Explanation:


Initial condition:

Balancing length \(l_1 = 40 cm\).

\[ \frac{P}{Q} = \frac{40}{60} \]
\[ \frac{4}{Q} = \frac{2}{3} \]
\[ Q = 6 \, \Omega \]

Second condition (after connecting \(x\) in series with \(P\)):

New resistance on left side \(= P + x = 4 + x\).

New balancing length \(l_2 = 80 cm\).

\[ \frac{4 + x}{Q} = \frac{80}{20} \]

Substitute \(Q = 6 \, \Omega\): \[ \frac{4 + x}{6} = 4 \]
\[ 4 + x = 24 \]
\[ x = 20 \, \Omega \]

Step 4: Final Answer:

The value of the unknown resistance is: \[ \boxed{20 \, \Omega} \] Quick Tip: In meter bridge problems, always apply the balance condition \(\frac{P}{Q} = \frac{l}{100-l}\).
When a resistance is added in series, simply add it algebraically before applying the ratio again.
Always measure the balancing length from the same end to avoid sign errors.

Step 1: Understanding the Question:

The circuit contains resistors, inductors, and capacitors connected in a complex network.

We are asked to find the effective current when the frequency is very high.

Hence, we must first simplify the circuit using high-frequency behavior of circuit elements.


Step 2: Key Formula or Approach:

At very high frequency:

Inductor reactance: \(X_L = \omega L \rightarrow \infty\) \(\Rightarrow\) Inductors behave as open circuits.

Capacitor reactance: \(X_C = \frac{1}{\omega C} \rightarrow 0\) \(\Rightarrow\) Capacitors behave as short circuits.


Step 3: Detailed Explanation:

Applying high-frequency conditions:


All inductors (\(20 mH, 50 mH\)) act as open circuits, so branches containing them are broken.

All capacitors (\(0.5 \muF, 0.8 \muF, 6 \muF\)) act as short circuits, directly connecting the nodes.


Due to capacitive shorting, the upper and lower horizontal branches become connected at the vertical capacitor positions.

After removing open branches and shorting capacitors, the circuit reduces to a purely resistive network.


The effective resistances remaining in the conducting path combine to give an equivalent resistance: \[ R_{eq} = 11 \, \Omega \]

Now applying Ohm's law using rms values: \[ I = \frac{V}{R_{eq}} = \frac{220}{11} \]
\[ I = 20 A \]

Step 4: Final Answer:

The effective current at very high frequency is: \[ \boxed{20 A} \] Quick Tip: For very high frequency AC circuits:
Inductors behave as open circuits and capacitors behave as short circuits.
Always simplify the circuit first before applying Ohm's law.
Such questions test conceptual understanding more than calculations.

Step 1: Understanding the Question:

A graph is plotted between \(\frac{1}{u}\) (x-axis) and \(\frac{1}{v}\) (y-axis) for a thin convex lens.

From the graph, we can determine the focal length of the lens.

Using the focal length and lens maker’s formula, we are asked to find the radius of curvature \(R\) of the lens.


Step 2: Key Formula or Approach:

Lens formula: \[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \]

For a graph of \(\frac{1}{v}\) versus \(\frac{1}{u}\), the y-intercept gives: \[ \frac{1}{f} \]

Lens maker’s formula for a thin symmetric convex lens: \[ \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

Step 3: Detailed Explanation:

From the graph:

Y-intercept \(= 0.10 \, cm^{-1}\).

\[ \frac{1}{f} = 0.10 \]
\[ f = 10 cm \]

Given: \(\mu = 1.5\).

For a symmetric convex lens: \[ R_1 = R, \quad R_2 = -R \]

Substitute in lens maker’s formula: \[ \frac{1}{f} = (1.5 - 1)\left(\frac{1}{R} - \left(-\frac{1}{R}\right)\right) \]
\[ \frac{1}{f} = 0.5 \cdot \frac{2}{R} = \frac{1}{R} \]

Thus: \[ R = f = 10 cm \]

Step 4: Final Answer:

The radius of curvature of each surface of the lens is: \[ \boxed{10 cm} \] Quick Tip: For a \(\frac{1}{v}\) vs \(\frac{1}{u}\) graph, the y-intercept directly gives \(\frac{1}{f}\).
Always remember: for a symmetric convex lens, \(R_1 = R\) and \(R_2 = -R\).
This type of question frequently appears in ray optics numerical sections.

Using the Rydberg formula \[ \frac{1}{\lambda} = R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \]

First Lyman line \((n_1=1,n_2=2)\): \[ \frac{1}{\lambda} = R\left(1-\frac14\right)=\frac{3R}{4} \Rightarrow \lambda=\frac{4}{3R} \]

3rd Paschen line \((n_1=3,n_2=6)\): \[ \frac{1}{\lambda_P} = R\left(\frac19-\frac1{36}\right)=\frac{R}{12} \Rightarrow \lambda_P=\frac{12}{R} \]

2nd Balmer line \((n_1=2,n_2=4)\): \[ \frac{1}{\lambda_B} = R\left(\frac14-\frac1{16}\right)=\frac{3R}{16} \Rightarrow \lambda_B=\frac{16}{3R} \]

Difference: \[ \Delta\lambda=\frac{12}{R}-\frac{16}{3R}=\frac{20}{3R} \]

Given \(\Delta\lambda=a\lambda\): \[ \frac{20}{3R}=a\left(\frac{4}{3R}\right)\Rightarrow a=5 \] Quick Tip: Always express unknown wavelengths as multiples of a known reference wavelength to eliminate the Rydberg constant quickly.

Maximum zener current occurs at maximum input voltage \(V_{in}=120\) V.

Total current: \[ I=\frac{V_{in}-V_z}{R_s}=\frac{120-60}{4\,k\Omega}=15~mA \]

Load current: \[ I_L=\frac{V_z}{R_L}=\frac{60}{10\,k\Omega}=6~mA \]

Zener current: \[ I_z=I-I_L=15-6=9~mA \] Quick Tip: For Zener regulators, always calculate \(I_z\) at maximum supply voltage to find the maximum safe current.

\(BrF_5\): The central atom Br has 7 valence electrons.\
It forms 5 \(\sigma\)-bonds with F atoms, using 5 pairs, and has 1 lone pair left.\
So, total electron pairs = 6 (5 bond pairs + 1 lone pair) \(\Rightarrow\) electron pair geometry is octahedral.\
Due to one lone pair, the actual \emph{molecular geometry becomes square pyramidal.

\([CrF_6]^{3-}\): Here Cr is in \(+3\) oxidation state and is surrounded by 6 F\(^{-}\) ligands.\
Coordination number = 6 implies an octahedral arrangement of ligands in most typical complexes.\
Therefore, the geometry of \([CrF_6]^{3-}\) is octahedral.

\(O_3\) (ozone): Ozone has a central O atom bonded to two terminal O atoms and possesses a lone pair on the central atom.\
Thus, there are 3 regions of electron density (2 bond pairs + 1 lone pair), leading to a bent (angular) molecular shape.\
\(PCl_5\): Phosphorus has 5 valence electrons and forms 5 \(\sigma\)-bonds with Cl without any lone pair on P in this molecule.\
According to VSEPR, 5 bonding pairs around the central atom give a trigonal bipyramidal geometry.


Hence: \(BrF_5 \rightarrow\) square pyramidal (II), \([CrF_6]^{3-} \rightarrow\) octahedral (IV), \(O_3 \rightarrow\) bent (I), \(PCl_5 \rightarrow\) trigonal bipyramidal (III).\ Quick Tip: Always first count total electron pairs (bond pairs + lone pairs) to get the electron pair geometry, then remove lone pairs to get the actual molecular shape.\ Square pyramidal comes from 6 pairs with 1 lone pair; trigonal bipyramidal comes from 5 bond pairs with no lone pair.\

Contact process is used for manufacture of \(H_2SO_4\) and the key step is catalytic oxidation of \(SO_2\) to \(SO_3\).\
The industrial catalyst for this step is \(V_2O_5\) supported on silica, so A \(\rightarrow III\).


Ostwald process is used for conversion of ammonia to nitric acid via oxidation of \(NH_3\) to \(NO\).\
This oxidation is carried out over a Pt–Rh gauze catalyst, hence B \(\rightarrow II\).


Haber process is used for synthesis of ammonia from \(N_2\) and \(H_2\).\
The industrial catalyst is finely divided iron (promoted with \(K_2O\), \(Al_2O_3\), etc.), therefore C \(\rightarrow I\).


Hydrogenation of oils (conversion of unsaturated vegetable oils to saturated fats) requires a metal catalyst.\
The commonly used catalyst is nickel (Raney Ni), so D \(\rightarrow IV\).\

Thus the correct matching is A-III, B-II, C-I, D-IV.\ Quick Tip: Remember processes along with their main products: Contact \(\rightarrow H_2SO_4\), Ostwald \(\rightarrow HNO_3\), Haber \(\rightarrow NH_3\), Hydrogenation of oils \(\rightarrow\) Vanaspati ghee.\ Then link each process to its characteristic catalyst: \(V_2O_5\), Pt–Rh, Fe, Ni in that order.\

\

For a homonuclear diatomic molecule like \(Cl_2\), the covalent radius is defined as half of the internuclear distance between the two nuclei.\
So, covalent radius \(r_cov = \dfrac{1}{2}(Cl–Cl bond length)\), not double the atomic radius, hence statement I is incorrect.\

When an atom gains one or more electrons to form an anion, the number of electrons increases while nuclear charge remains the same.\
This increases electron–electron repulsion and effectively reduces nuclear attraction per electron, leading to expansion of the electron cloud.\
Therefore, anionic radius is larger than the radius of the corresponding neutral atom, so statement II is correct.\ Quick Tip: For \(X_2\) molecules, always think: bond length = \(2 \times\) covalent radius.\ For isoelectronic species, more negative charge \(\Rightarrow\) larger radius; more positive charge \(\Rightarrow\) smaller radius.\

Liquation is a refining method in which an impure metal is gently heated to a temperature just above its melting point.\
The metal with relatively low melting point melts and flows away, while high-melting solid impurities remain behind.


Because the process relies on selective melting, it is suitable only for metals that can be melted without melting their impurities, i.e., metals having comparatively low melting points (like Sn, Pb, Bi).\
Hence, metals with low melting points are best refined by liquation.\ Quick Tip: Think: “Liquation” \(\rightarrow\) “Liquid metal flows away”.\ Whenever a question mentions a soft, low-melting metal (Sn, Pb, Bi), liquation is a likely refining method.\

\

Hydrogen peroxide decomposes slowly to \(H_2O\) and \(O_2\), and the decomposition is catalysed by light, heat, and traces of metal ions.\
To slow this decomposition, stabilizers (negative catalysts) like urea, phosphoric acid, or acetanilide are added to absorb impurities or form complexes with metal ions.


Urea is a commonly used stabilizer that reduces the rate of decomposition by tying up catalytic impurities, thus preserving \(H_2O_2\) for longer time.\
Therefore, urea prevents decomposition of hydrogen peroxide.\ Quick Tip: Remember: acetanilide and urea are classic stabilizers of \(H_2O_2\) used in laboratory preparations.\ Stabilizer \(\Rightarrow\) negative catalyst: it lowers the rate of an otherwise spontaneous decomposition.\

Lithium aluminium hydride (\(LiAlH_4\)) is a strong hydride donor and a powerful reducing agent.

When \(BeCl_2\) reacts with \(LiAlH_4\), hydride ions (\(H^-\)) from \(AlH_4^-\) replace chloride ions of beryllium, forming beryllium hydride.


The balanced reaction is:

2
B
e
C
l
2
+
L
i
A
l
H
4
→
2
B
e
H
2
+
L
i
C
l
+
A
l
C
l
3
2BeCl
2
+LiAlH
4
→2BeH
2
+LiCl+AlCl
3


Here, \(BeH_2\) is formed as the main hydride product, while \(LiCl\) and \(AlCl_3\) are formed as by-products due to exchange of \(Cl^-\) between \(BeCl_2\) and \(LiAlH_4\).\ Quick Tip: \(LiAlH_4\) converts metal halides (especially of \(s\)-block and some \(p\)-block metals) into corresponding hydrides by hydride transfer.\ Check charge balance and total halide ions to quickly write balanced equations in such conversions.\

Borazine (\(B_3N_3H_6\)) is obtained by heating diborane with excess ammonia.\
The overall reaction can be written as:


3
B
2
H
6
+
6
N
H
3
→
2
B
3
N
3
H
6
+
12
H
2
3B
2
H
6
+6NH
3
→2B
3
N
3
H
6
+12H
2


Here, X = \(B_2H_6\) (diborane) and Y = \(NH_3\) (ammonia).

In this process, B–H and N–H bonds rearrange to form B–N ring structure with release of hydrogen gas.\
Borazine has an alternating B and N ring, analogous in structure to benzene, and hence is called “inorganic benzene”.\ Quick Tip: Link the name “borazine” with “boron + azine (nitrogen)” to recall that it comes from \(B_2H_6\) and \(NH_3\).\ Inorganic benzene \(\Rightarrow\) B\(_3\)N\(_3\) ring, prepared from diborane and ammonia on heating.\

In disproportionation, the same element in a single oxidation state is simultaneously oxidized and reduced to two different oxidation states.


(A) In \(2H_2O_2 \rightarrow 2H_2O + O_2\), O in \(H_2O_2\) is in \(-1\) state.\
It goes to \(-2\) in \(H_2O\) (reduction) and \(0\) in \(O_2\) (oxidation), so O undergoes both oxidation and reduction \(\Rightarrow\) disproportionation.


(B) In \(2NO_2 + H_2O \rightarrow HNO_3 + HNO_2\), N in \(NO_2\) is at \(+4\) oxidation state.\
In \(HNO_3\), N is \(+5\) (oxidation) and in \(HNO_2\), N is \(+3\) (reduction), so this is also a disproportionation of nitrogen.


(C) \(MnO_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O\) is a reduction half-reaction: Mn changes from \(+7\) (in \(MnO_4^-\)) to \(+4\) (in \(MnO_2\)).\
Only reduction is taking place; there is no simultaneous oxidation of Mn in this single step, so it is not a disproportionation reaction.


(D) In \(3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2\), Mn is initially in \(+6\) (in \(MnO_4^{2-}\)).

It changes to \(+7\) in \(MnO_4^-\) (oxidation) and to \(+4\) in \(MnO_2\) (reduction), so Mn is both oxidized and reduced \(\Rightarrow\) disproportionation.


Thus, only (C) is not a disproportionation reaction.\ Quick Tip: For quick checking, write oxidation numbers of the central element on both sides.\ If a single species of an element gives two different oxidation states, it is disproportionation; if it goes only up or only down, it is simple oxidation or reduction.\

In acidic medium, \(KMnO_4\) acts as a strong oxidizing agent.

The relevant half-reaction for \(MnO_4^-\) in acidic solution is:


M
n
O
4
−
+
8
H
+
+
5
e
−
→
M
n
2
+
+
4
H
2
O
MnO
4
−
+8H
+
+5e
−
→Mn
2+
+4H
2
O

In \(MnO_4^-\), Mn is in oxidation state \(+7\).

In \(Mn^{2+}\), Mn is in oxidation state \(+2\).

Thus, the change in oxidation number of Mn is \(7 - 2 = 5\) (reduction by 5 units).


Therefore, the change in oxidation number is 5.\ Quick Tip: Remember typical \(KMnO_4\) reductions: acidic medium \(\Rightarrow Mn^{2+}\) (change of 5), neutral/basic medium \(\Rightarrow MnO_2\) (change of 3).\ This directly gives the \(n\)-factor for calculating equivalents in redox titrations.\

The dot over Cl indicates a chlorine free radical (\(\cdot Cl\)).

In the free radical halogenation of alkanes, the first propagation step is abstraction of a hydrogen atom from the alkane by the halogen radical.


Here, \(\cdot Cl\) abstracts a hydrogen atom from methane:


⋅
C
l
+
C
H
4
→
H
C
l
+
⋅
C
H
3
⋅Cl+CH
4
→HCl+⋅CH
3

So, the products are HCl (a stable molecule) and the methyl radical (\(\cdot CH_3\)).\
Hence, A is \(CH_3\) (radical) and B is HCl.\ Quick Tip: In radical halogenation of alkanes: halogen radical + alkane \(\rightarrow\) HX + alkyl radical is the key propagation step.\ Remember that radical reactions generally maintain one radical on both sides to continue the chain process.\

The sample size is about 100 mg, which is in the milligram range.\
For such small amounts, preparative thin layer chromatography (preparative TLC) is an efficient and economical method to separate closely related organic compounds.


p-Nitrophenol and picric acid have similar aromatic structures but differ in the number of nitro groups and acidity.

Their \(R_f\) values on a suitable TLC plate will be different enough to allow separation.

Preparative TLC allows you to run a thicker layer, separate the bands, and scrape off and elute each compound.


Column chromatography is more suitable for gram-level separations, while preparative TLC is ideal for tens to hundreds of milligrams.
Quick Tip: As a thumb rule: \(\sim\)mg scale \(\Rightarrow\) preparative TLC; multi-gram scale \(\Rightarrow\) column chromatography.\ When two compounds are similar but show different \(R_f\) on TLC, scaling up to preparative TLC is often the fastest lab choice.\

Phenol undergoes electrophilic bromination.\
In a non-polar solvent like \(CCl_4\) or \(CHCl_3\), phenol remains largely in its neutral molecular form.

Therefore, its activating –OH group increases reactivity moderately, giving predominantly mono-bromophenol (usually at the para or ortho position).


In polar protic solvents like water, phenol is partially ionized to phenoxide ion (\(C_6H_5O^-\)), which is a much stronger activating group than –OH.


Phenoxide ion greatly increases the electron density at ortho and para positions, leading to rapid and multiple substitution.\
Thus, aqueous bromine with phenol quickly gives 2,4,6-tribromophenol (white precipitate).


The key difference is the polarity and protic nature of the solvent, which controls the extent of ionization of phenol and hence its reactivity towards bromination.
Quick Tip: Remember: phenol + \(Br_2\) in \(CCl_4\) \(\rightarrow\) mono-bromo; phenol + \(Br_2\) in water \(\rightarrow\) 2,4,6-tribromophenol.\ Polar protic solvent (water) enhances formation of phenoxide, making the ring much more reactive towards multiple electrophilic substitutions.\

Step 1: Understanding the Question:


A compound is aromatic only if it is cyclic, planar, fully conjugated and obeys Hückel’s rule, i.e. it must contain \((4n+2)\pi\) electrons.


Step 2: Key Rule:
\[ Aromatic \Rightarrow Cyclic + Planar + Fully conjugated + (4n+2)\pi electrons \]

Step 3: Detailed Explanation:


- Option (A): Benzene has 6\(\pi\) electrons and satisfies all aromatic conditions.

- Option (B): Cyclopentadienyl anion has 6\(\pi\) electrons and is aromatic.

- Option (C): Tropylium cation is planar and has 6\(\pi\) electrons, hence aromatic.

- Option (D): Cyclopentadiene contains one \(sp^3\) hybridized carbon which breaks conjugation, making it non-aromatic.


Step 4: Final Answer:


Cyclopentadiene is not aromatic.
Quick Tip:
Check for \(sp^3\) carbons first!
If a ring has an \(sp^3\) hybridized atom in the ring itself, the system cannot be fully conjugated and is typically non-aromatic, regardless of the electron count.

Step 1: Understanding the Question:


The reaction involves an aldehyde group along with a phenolic group subjected to Tollens’ reagent and sodium borohydride.


Step 2: Key Reactions:


- Tollens’ reagent oxidizes aldehydes to carboxylate ions.

- \(NaBH_4\) reduces aldehydes to alcohols but does not reduce carboxylates.


Step 3: Detailed Explanation:


- With Tollens’ reagent, the aldehyde group is oxidized to the carboxylate ion, forming 4-hydroxybenzoate ion (A).


- With \(NaBH_4\), the aldehyde group is reduced to alcohol, forming 4-hydroxybenzyl alcohol (B).


Step 4: Final Answer:
\(A =\) 4-hydroxybenzoate ion and \(B =\) 4-hydroxybenzyl alcohol.
Quick Tip: Tollen's reagent is a mild oxidizing agent specific to aldehydes.
\(NaBH_4\) is a selective reducing agent that reduces aldehydes and ketones but is too weak to reduce carboxylic acids or esters.

Step 1: Understanding the Question:

The product formed is a tertiary alcohol containing two phenyl groups and one methyl group.


Step 2: Key Concept:

Grignard reagents react with ketones using one equivalent to form tertiary alcohols.


Step 3: Detailed Explanation:

- Acetophenone (\(Ph-CO-CH_3\)) reacts with one mole of \(PhMgBr\) to give a tertiary alcohol after hydrolysis.

- Nitriles and esters require two equivalents of Grignard reagent to form tertiary alcohols.


Step 4: Final Answer:
\(Ph-CO-CH_3\) is the correct reactant.
Quick Tip: To find the starting materials for a tertiary alcohol \(R_1R_2R_3C-OH\):
Break one C-R bond. The R group comes from the Grignard (\(R-MgX\)) and the remaining part (\(R_1-CO-R_2\)) is the starting ketone.

Step 1: Understanding the Question:


The reaction conditions given are Na in liquid ammonia followed by alcohol.

This is the characteristic condition for Birch reduction.

We must predict the major product based on the nature of substituents present on the benzene ring.


Step 2: Key Formula or Approach:


Rules for Birch reduction of substituted benzene rings:

Electron-donating groups (EDG) such as \(-\mathrm{OCH_3}\) keep their carbon unsaturated (\(sp^2\)).

Electron-withdrawing groups (EWG) such as \(-\mathrm{NO_2}\) cause their carbon to become saturated (\(sp^3\)).


Step 3: Detailed Explanation:


The given aromatic ring contains:
\(-\mathrm{OCH_3}\): an electron-donating group (+M effect).
\(-\mathrm{NO_2}\): a strong electron-withdrawing group (–M and –I effects).


During Birch reduction:

Two electrons are added to the aromatic ring followed by protonation.

Reduction does not occur uniformly across the ring.


Effect of substituents:

The carbon attached to \(-\mathrm{OCH_3}\) resists reduction and remains \(sp^2\)-hybridized.

The carbon attached to \(-\mathrm{NO_2}\) undergoes reduction and becomes \(sp^3\)-hybridized.


Hence, the final product is a 1,4-cyclohexadiene derivative in which:

Double bonds are located away from the \(-\mathrm{OCH_3}\) group.

The \(-\mathrm{NO_2}\)-bearing carbon is saturated.


Step 4: Final Answer:


The major product is the Birch reduction product in which the methoxy-substituted carbon remains unsaturated and the nitro-substituted carbon is reduced.
Quick Tip: In Birch reduction:
Electron-donating groups keep their attached carbon \(sp^2\).
Electron-withdrawing groups force their attached carbon to become \(sp^3\).
Always identify EDG vs EWG before predicting the product.

Step 1: Understanding the Question:

We are asked to compare the relative stability of aromatic diazonium salts.

The stability of diazonium salts depends strongly on the nature of substituents present on the benzene ring.


Step 2: Key Formula or Approach:

Diazonium ion stability increases with electron-donating groups (EDG) on the aromatic ring.

Electron-withdrawing groups (EWG) destabilize the diazonium ion.


Step 3: Detailed Explanation:

Let us analyze each substituent:


(A) \(-\mathrm{OCH_3}\):

Strong electron-donating group due to \(+M\) (resonance) effect.

Stabilizes the diazonium ion by delocalizing positive charge.


(C) No substituent:

Neutral benzene diazonium ion.

Moderate stability, lower than EDG substituted ring.


(D) \(-\mathrm{CN}\):

Strong electron-withdrawing group due to \(-M\) and \(-I\) effects.

Destabilizes the diazonium ion.


(B) \(-\mathrm{NO_2}\):

Very strong electron-withdrawing group with both \(-M\) and \(-I\) effects.

Greatly destabilizes the diazonium ion, making it least stable.


Thus, the correct decreasing order of stability is: \[ (A) > (C) > (D) > (B) \]

Step 4: Final Answer:

The correct stability order is: \[ \boxed{(A) > (C) > (D) > (B)} \] Quick Tip: Diazonium ions are stabilized by electron-donating groups through resonance.
Strong electron-withdrawing groups drastically reduce stability.
Always rank substituents using their \(+M/-M\) effects for such questions.

Step 1: Understanding the Question:

Stearic acid reacts with polyethylene glycol to form an ester.


Step 2: Key Concept:

Non-ionic detergents do not carry any ionic charge on the hydrophilic head.


Step 3: Detailed Explanation:

The product contains a long hydrophobic hydrocarbon chain and a neutral polyether hydrophilic head. Hence, it is a non-ionic detergent.


Step 4: Final Answer:

The product is a non-ionic detergent.
Quick Tip: Mnemonic for detergents:
- Sulfonates/Sulfates \(\rightarrow\) Anionic (e.g., Sodium lauryl sulfate).
- Quaternary Ammonium salts \(\rightarrow\) Cationic.
- Polyethylene glycol esters \(\rightarrow\) Non-ionic.

Step 1: Understanding the Question:

Reducing sugars possess a free hemiacetal or hemiketal group that can open to form an aldehyde or ketone.


Step 2: Key Identification:

Presence of a free anomeric carbon bonded to an –OH group indicates a reducing sugar.


Step 3: Detailed Explanation:

- Options with acetal or glycosidic linkages at the anomeric carbon are non-reducing.

- Option (C) has a free hemiacetal group, making it capable of reducing Tollens’ or Fehling’s reagent.


Step 4: Final Answer:

Option (C) is a reducing sugar.
Quick Tip: Look at the anomeric carbon (the one attached to the ring oxygen and another oxygen).
If it is attached to -OH \(\rightarrow\) Reducing Sugar.
If it is attached to -OCH\(_3\) or another sugar ring \(\rightarrow\) Non-reducing Sugar.

Step 1: Understanding the Question:

The question tests the behavior of aniline during Friedel--Crafts alkylation.

We must analyze whether methylation occurs at ortho/para positions or not, and why.


Step 2: Key Formula or Approach:

Friedel--Crafts alkylation requires an activated aromatic ring.

Lewis acids like anhydrous \( \mathrm{AlCl_3} \) can interact strongly with basic groups such as \( -\mathrm{NH_2} \).


Step 3: Detailed Explanation:

Aniline normally has a strongly activating \( -\mathrm{NH_2} \) group, which is ortho/para directing.

However, in the presence of anhydrous \( \mathrm{AlCl_3} \), the lone pair on nitrogen coordinates with \( \mathrm{AlCl_3} \) forming a complex or salt: \[ \mathrm{C_6H_5NH_2 + AlCl_3 \rightarrow C_6H_5NH_2 \cdot AlCl_3} \]
This coordination removes the electron-donating ability of the \( -\mathrm{NH_2} \) group.

As a result, the benzene ring becomes deactivated toward electrophilic substitution.

Due to this deactivation, the usual ortho/para directing influence is lost.

Under these conditions, substitution does not give \( o \)- or \( p \)-methylaniline experimentally.


Hence:

Assertion (A) is true because ortho and para products are not formed.

Reason (R) is also true because salt formation with \( \mathrm{AlCl_3} \) deactivates the ring and explains the observation.


Step 4: Final Answer:

Both the Assertion and Reason are true, and the Reason correctly explains the Assertion.
Quick Tip: In Friedel--Crafts reactions, strongly basic groups like \( -\mathrm{NH_2} \) form complexes with \( \mathrm{AlCl_3} \).
Such complex formation deactivates the aromatic ring and alters the usual directing effects.
Always check Lewis acid--base interactions before predicting products.

Step 1: Understanding the Question:

We are given concentration of Mg in ppm and total volume. We must find the total number of Mg atoms.


Step 2: Key Formula:
\(ppm = mg L^{-1}\) (for dilute aqueous solutions)

Number of atoms \(= \dfrac{mass}{molar mass} \times N_A\)


Step 3: Calculation:

Mass of Mg \(= 48 \times 2 = 96\) mg \(= 96 \times 10^{-3}\) g

Moles of Mg \(= \dfrac{96 \times 10^{-3}}{24} = 4 \times 10^{-3}\) mol

Atoms \(= 4 \times 10^{-3} \times 6.02 \times 10^{23} \approx 24 \times 10^{20}\)


Step 4: Final Answer:
\(x = 24\)
Quick Tip: 1 ppm \(\approx\) 1 mg L\(^{-1}\) for dilute solutions. Always convert mg to grams before mole calculation.

Step 1: Understanding the Question:

The mixture composition is given in terms of mass percentage.

To calculate partial pressure, mole fraction must be used.


Step 2: Key Formula or Approach:

Partial pressure \(=\) Mole fraction \(\times\) Total pressure.


Step 3: Detailed Explanation:

Assume total mass of mixture \(= 100\) g.


Mass of \(H_2 = 40\) g \(\Rightarrow\) moles of \(H_2 = \dfrac{40}{2} = 20\).

Mass of \(O_2 = 60\) g \(\Rightarrow\) moles of \(O_2 = \dfrac{60}{32} = 1.875\).


Total moles \(= 20 + 1.875 = 21.875\).


Mole fraction of \(H_2\): \[ X_{H_2} = \frac{20}{21.875} \approx 0.914 \]

Partial pressure of \(H_2\): \[ P_{H_2} = 0.914 \times 2.2 \approx 2.0~bar \]

Step 4: Final Answer:

Partial pressure of hydrogen \(= 2.0\) bar.
Quick Tip: Partial pressure depends on mole fraction, not mass fraction.

Step 1: Understanding the Question:

The condition for equal wavelengths of particles is applied.


Step 2: Key Formula or Approach:

de Broglie relation: \[ \lambda = \frac{h}{mv} \]

Step 3: Detailed Explanation:

For equal wavelengths: \[ m_e v_e = m_n v_n \]
\[ \frac{v_e}{v_n} = \frac{m_n}{m_e} \]

Substitute values: \[ x = \frac{1.6 \times 10^{-27}}{9.1 \times 10^{-31}} \approx 1758 \]

Step 4: Final Answer:
\(x = 1758\).
Quick Tip: Equal wavelength \(\Rightarrow\) equal momentum.
Velocity is inversely proportional to mass.

Step 1: Understanding the Question:

Combustion occurs at constant volume using a bomb calorimeter.


Step 2: Key Formula or Approach:

Heat absorbed: \[ q = C \Delta T \]

Step 3: Detailed Explanation:

Temperature rise: \[ \Delta T = 300 - 298 = 2~K \]
\[ q = 20 \times 2 = 40~kJ \]

Moles of carbon: \[ n = \frac{2.4}{12} = 0.2 \]

Enthalpy change per mole: \[ \Delta H = \frac{40}{0.2} = 200~kJ mol^{-1} \]

Step 4: Final Answer:
\(x = 200\).
Quick Tip: For solid combustion, \(\Delta n_g = 0\), hence \(\Delta H \approx \Delta U\).

Step 1: Understanding the Question:

Heating causes volume reduction and loss of solute.


Step 2: Key Formula or Approach:

Molarity \(=\) moles of solute / volume of solution.


Step 3: Detailed Explanation:

Initial moles: \[ 0.5 \times 0.8 = 0.4 \]

Evaporated moles: \[ \frac{11.5}{63} \approx 0.1825 \]

Remaining moles: \[ 0.4 - 0.1825 = 0.2175 \]

Final volume: \[ \frac{800}{2} = 400~mL = 0.4~L \]

Final molarity: \[ \frac{0.2175}{0.4} = 0.54375 \approx 54 \times 10^{-2} \]

Step 4: Final Answer:
\(x = 54\).
Quick Tip: If solute loss is mentioned explicitly, subtract it before recalculating molarity.

Step 1: Understanding the Question:

The reaction is reversed and multiplied by \(\frac{1}{2}\).


Step 2: Key Formula or Approach:

Reverse reaction \(\Rightarrow 1/K\).

Multiplying equation by \(n\) \(\Rightarrow K^n\).


Step 3: Detailed Explanation:
\[ K' = (2 \times 10^{15})^{-1/2} \]
\[ K' \approx 2 \times 10^{-8} \]

Step 4: Final Answer:
\(x = 2\).
Quick Tip: Reverse \(\Rightarrow 1/K\), multiply equation by \(n\) \(\Rightarrow K^n\).

Step 1: Understanding the Question:

Iron exists in mixed oxidation states in \(Fe_3O_4\).


Step 3: Detailed Explanation:
\(Fe_3O_4 = FeO \cdot Fe_2O_3\).


Average oxidation state: \[ \frac{2 + 6}{3} = \frac{8}{3} \]

Electrons required per iron atom \(= \frac{8}{3} \approx 3\).


Step 4: Final Answer:

Required charge \(= 3\) Faraday.
Quick Tip: \(Fe_3O_4\) is a mixed oxide: \(FeO \cdot Fe_2O_3\).

Step 1: Understanding the Question:

Half-life changes with concentration.


Step 2: Key Formula or Approach:
\[ t_{1/2} \propto [A]^{1-n} \]

Step 3: Detailed Explanation:
\[ \frac{100}{50} = (0.5)^{1-n} \]
\[ 2 = 2^{n-1} \Rightarrow n = 2 \]

Step 4: Final Answer:

Order of reaction \(= 2\).
Quick Tip: Doubling concentration halves half-life \(\Rightarrow\) second-order reaction.

Step 1: Understanding the Question:

Magnetic moment depends on number of unpaired electrons.


Step 3: Detailed Explanation:

Both complexes contain 3 unpaired electrons.

\[ \mu = \sqrt{n(n+2)} \]

Thus, magnetic moments are equal.


Step 4: Final Answer:

Difference \(= 0\).
Quick Tip: Same number of unpaired electrons \(\Rightarrow\) same magnetic moment.

Step 1: Understanding the Question:

Reaction conditions determine the reaction type.


Step 3: Detailed Explanation:

In sunlight, benzene undergoes addition reaction with chlorine.

Product formed is benzene hexachloride (BHC): \(C_6H_6Cl_6\).


Number of hydrogen atoms remains unchanged.


Step 4: Final Answer:

Number of hydrogens \(= 6\).
Quick Tip: \(Cl_2\)/Sunlight \(\Rightarrow\) Addition (BHC).
\(Cl_2\)/FeCl\(_3\) \(\Rightarrow\) Substitution.