JEE Physics Mathematical Tools MCQs Set 04

Answer: (b) \( \dfrac{3\pi^2}2 \)

Explanation:
1. \( \left[\dfrac{\theta^2}2\right]_\pi^{2\pi}=\dfrac{4\pi^2-\pi^2}2 \).
2. \( =\dfrac{3\pi^2}2 \).

Teacher's Note:
1. Subtract the lower-limit value.
2. \( (2\pi)^2=4\pi^2 \).

Answer: (a) 7/3

Explanation:
1. \( \left[\dfrac{x^3}3\right]_0^{7^{1/3}}=\dfrac73 \).

Teacher's Note:
1. \( (7^{1/3})^3=7 \).
2. The lower limit contributes 0.

Answer: (d) 1

Explanation:
1. Put \( u=x^2 \), \( x\,dx=du/2 \); the limits become 0 and π.
2. \( \dfrac12\int_0^\pi\sin u\,du=\dfrac12\cdot2=1 \).

Teacher's Note:
1. Change the limits along with the variable.
2. \( \int_0^\pi\sin u\,du=2 \).

Answer: (c) \( \dfrac13\ln\dfrac52 \)

Explanation:
1. \( \left[\dfrac13\ln(3x+2)\right]_0^1=\dfrac13(\ln5-\ln2) \).
2. \( =\dfrac13\ln\dfrac52 \).

Teacher's Note:
1. Divide by the coefficient of x.
2. Subtract logarithms for the ratio.

Answer: (b) \( e-1 \)

Explanation:
1. \( [e^x]_0^1=e-1 \).

Teacher's Note:
1. The lower limit gives \( e^0=1 \).
2. The result is about 1.718.

Answer: (a) \( b^2 \)

Explanation:
1. Area = \( \int_0^b2x\,dx=[x^2]_0^b=b^2 \).

Teacher's Note:
1. It is also the area of a triangle: \( \tfrac12\cdot b\cdot2b \).
2. Both methods agree.

Answer: (d) \( \dfrac{b(4+b)}4 \)

Explanation:
1. \( \int_0^b\left(\dfrac x2+1\right)dx=\dfrac{b^2}4+b \).
2. \( =\dfrac{b(b+4)}4 \).

Teacher's Note:
1. Integrate each term.
2. It is a trapezium with parallel sides 1 and \( 1+b/2 \).

Answer: (c) 2 square units

Explanation:
1. \( \int_0^\pi\sin x\,dx=[-\cos x]_0^\pi=1+1=2 \).

Teacher's Note:
1. \( \sin x\ge0 \) on [0, π], so no sign problem.
2. Area under one arch of sine is 2.

Answer: (b) \( b^3 \)

Explanation:
1. \( \int_0^b3x^2\,dx=[x^3]_0^b=b^3 \).

Teacher's Note:
1. The coefficient 3 cancels the 1/3 from integration.
2. Check: derivative of \( b^3 \) is \( 3b^2 \).

Answer: (a) I-r, II-p, III-q, IV-s

Explanation:
1. \( \int\sec x\tan x\,dx=\sec x \).
2. \( \int\csc Kx\cot Kx\,dx=-\dfrac{\csc Kx}K \) and \( \int\csc^2Kx\,dx=-\dfrac{\cot Kx}K \).
3. \( \int\cos Kx\,dx=\dfrac{\sin Kx}K \).

Teacher's Note:
1. Differentiate each candidate result to check.
2. The 1/K factor comes from the inner derivative.

Answer: (d) 30 m east

Explanation:
1. North and south legs cancel: \( 40-40=0 \).
2. Only the 30 m east displacement remains.

Teacher's Note:
1. Displacement depends only on the end points.
2. The distance walked is 110 m.

Answer: (c) 50, \( 53^\circ \) north of east

Explanation:
1. \( R=\sqrt{30^2+40^2}=50 \).
2. \( \tan\alpha=\dfrac{40}{30} \), so \( \alpha=53^\circ \) with the east direction.

Teacher's Note:
1. The angle with east is the larger one because the northward part is bigger.
2. 3-4-5 triangle.

Answer: (b) magnitude only

Explanation:
1. Interchanging does not change the angle between the vectors.
2. \( |\vec c|^2=a^2+b^2+2ab\cos\theta \) is unchanged, but the direction with respect to \( \vec a \) now differs (the resultant swings towards the other vector).

Teacher's Note:
1. Magnitude depends only on a, b and the angle between them.
2. Direction relative to a given vector is not symmetrical.

Answer: (a) cannot be zero

Explanation:
1. \( \vec A+\vec B \) lies in the plane, while \( \vec C \) has a component out of the plane.
2. So the sum has a non-zero out-of-plane component and cannot vanish.

Teacher's Note:
1. A vector with an out-of-plane part cannot be cancelled by in-plane vectors.
2. Think of the z component.

Answer: (d) 8 N

Explanation:
1. The resultant lies between \( 10-6=4 \) N and \( 10+6=16 \) N.
2. Only 8 N lies in the range.

Teacher's Note:
1. Use the triangle inequality.
2. The extremes occur at 180° and 0°.

Answer: (c) null vector

Explanation:
1. The tip of the last vector returns to the tail of the first.
2. The net displacement is zero, so the resultant is the null vector.

Teacher's Note:
1. A closed polygon means the vectors sum to zero.
2. This is the polygon law.

Answer: (b) \( \pi \)

Explanation:
1. \( R^2=P^2+Q^2+2PQ\cos\theta \) is smallest when \( \cos\theta=-1 \).
2. So \( \theta=\pi \) and \( R=|P-Q| \).

Teacher's Note:
1. Opposite directions give the smallest resultant.
2. The maximum is at θ = 0.

Answer: (a) \( \pi/2 \)

Explanation:
1. \( 13^2=12^2+5^2+2(12)(5)\cos\theta \).
2. \( 169=169+120\cos\theta \), so \( \cos\theta=0 \) and \( \theta=\pi/2 \).

Teacher's Note:
1. 12-5-13 is a Pythagorean triple.
2. A right angle between the vectors gives \( C^2=A^2+B^2 \).

Answer: (d) also perpendicular to each other and of equal lengths

Explanation:
1. Take \( \vec A=a\hat i \), \( \vec B=a\hat j \): sum \( a(\hat i+\hat j) \), difference \( a(\hat i-\hat j) \).
2. Each has length \( a\sqrt2 \) and their dot product is \( a^2-a^2=0 \).

Teacher's Note:
1. Use components to test the angle.
2. The diagonals of a square are equal and perpendicular.

Answer: (c) \( \sqrt{14} \)

Explanation:
1. \( \sqrt{9+4+1}=\sqrt{14} \).

Teacher's Note:
1. Add the squares of the components.
2. Take the square root at the end.

Answer: (b) \( -25\cos30^\circ \) and \( -25\sin30^\circ \)

Explanation:
1. \( x=25\cos210^\circ=-25\cos30^\circ\approx-21.7 \) m.
2. \( y=25\sin210^\circ=-25\sin30^\circ=-12.5 \) m.

Teacher's Note:
1. 210° lies in the third quadrant where both components are negative.
2. \( \cos(180^\circ+\alpha)=-\cos\alpha \).

Answer: (a) \( 30\sqrt3 \) km/h

Explanation:
1. \( v_2=\sqrt{60^2-30^2}=\sqrt{2700} \).
2. \( =30\sqrt3\approx52 \) km/h.

Teacher's Note:
1. Components are perpendicular, so use Pythagoras.
2. The other component must be less than 60 km/h.

Answer: (d) \( \pm\dfrac{\sqrt{11}}{10} \)

Explanation:
1. \( 0.25+0.64+C^2=1 \).
2. \( C^2=0.11 \), so \( C=\pm\sqrt{0.11}=\pm\dfrac{\sqrt{11}}{10} \).

Teacher's Note:
1. The length must equal 1.
2. Both signs satisfy the condition.

Answer: (c) \( 15^\circ \)

Explanation:
1. The first makes \( \tan^{-1}(2/2)=45^\circ \) with the x-axis, the second \( \tan^{-1}\sqrt3=60^\circ \).
2. The angle between them is \( 60^\circ-45^\circ=15^\circ \).

Teacher's Note:
1. Find each direction angle.
2. Check with the dot product: cosθ = (2+2√3)/(2√2·2) = 0.966.

Answer: (b) \( 2\hat i-3\hat j \)

Explanation:
1. A force with x-component 2 N and y-component −3 N is \( 2\hat i-3\hat j \).

Teacher's Note:
1. Components keep their signs.
2. Do not swap the components.

Answer: (a) \( \hat i-4\hat j+5\hat k \)

Explanation:
1. \( \vec{AB}=\vec r_B-\vec r_A=(2-1)\hat i+(-3-1)\hat j+(4+1)\hat k \).
2. \( =\hat i-4\hat j+5\hat k \).

Teacher's Note:
1. Final minus initial.
2. Subtracting −1 adds 1 in the z component.