JEE Physics Basic Mathematics Used In Physics MCQs Set 06

Answer: (b) 2

Explanation:
1. Geometric series with a = 1, r = 1/2: \( S = \frac{a}{1-r} = 2 \).

Teacher's Note:
1. Use the infinite GP formula.
2. |r| < 1.

Answer: (a) \( \frac{\log_e(2x-3)}{2}+c \)

Explanation:
1. Let \( u = 2x-3 \), \( du = 2dx \). \( \int\frac{du}{2u} = \frac12\ln u + c \).

Teacher's Note:
1. Substitute for the linear expression.
2. ∫ du/u = ln u.

Answer: (c) \( 1-\frac{2h}{R} \)

Explanation:
1. Binomial approximation \( (1+x)^n \approx 1+nx \) for \( x\ll1 \), with n = −2 and \( x = \frac hR \).

Teacher's Note:
1. Drop terms in (h/R)².
2. Used often for g at small heights.

Answer: (a) 0.999

Explanation:
1. \( (1-0.004)^{1/4}\approx1-\frac{0.004}{4} = 0.999 \).

Teacher's Note:
1. Binomial approximation.
2. Fourth root changes the correction by 1/4.

Answer: (d) \( -\frac{\cos4x}{4}+c \)

Explanation:
1. \( \int\sin ax\,dx = -\frac{\cos ax}{a}+c \) with a = 4.

Teacher's Note:
1. Divide by the coefficient of x.
2. Integral of sin is −cos.

Answer: (b) \( \frac{\sqrt3}{2} \)

Explanation:
1. \( \sin480^\circ = \sin(360^\circ+120^\circ) = \sin120^\circ = \sin60^\circ = \frac{\sqrt3}2 \).

Teacher's Note:
1. Subtract 360°.
2. sin(180° − θ) = sin θ.

Answer: (c) \( \sin x+x\cos x \)

Explanation:
1. Product rule: \( \frac{dy}{dx} = 1\cdot\sin x + x\cos x \).

Teacher's Note:
1. (uv)' = u'v + uv'.
2. Differentiate each factor once.

Answer: (d) \( -\frac A2 \) and \( -\frac{\sqrt3A}{2} \)

Explanation:
1. \( A_x = A\cos240^\circ = -\frac A2 \), \( A_y = A\sin240^\circ = -\frac{\sqrt3}2A \).

Teacher's Note:
1. 240° lies in the third quadrant.
2. Both components are negative.

Answer: (b) 15°

Explanation:
1. First ball makes \( \tan^{-1}\sqrt3 = 60^\circ \) with the x-axis; second makes \( \tan^{-1}1 = 45^\circ \).
2. The angle between the paths is 60° − 45° = 15°.

Teacher's Note:
1. Find each direction from the x-axis.
2. Subtract the angles.

Answer: (a) \( \frac{\pi\sqrt2}{30} \) cm/s

Explanation:
1. Speed \( v = \omega r = \frac{2\pi}{60}\times1 = \frac\pi{30} \) cm/s.
2. In 15 s the hand turns 90°, so the velocity vectors are perpendicular: |Δv| = √2 v.

Teacher's Note:
1. A quarter turn rotates the velocity by 90°.
2. |Δv| = 2v sin(θ/2).

Answer: (c) \( |\vec A|=|\vec B| \)

Explanation:
1. The resultant bisects the angle between two vectors only when they have equal magnitude (the parallelogram is a rhombus).

Teacher's Note:
1. R along the diagonal of a rhombus bisects the angle.
2. tan α = B sin θ / (A + B cos θ).

Answer: (c) parallel

Explanation:
1. \( |\vec C|^2 = A^2+B^2+2AB\cos\theta = (A+B)^2 \Rightarrow \cos\theta = 1 \).
2. So θ = 0, the vectors are parallel.

Teacher's Note:
1. Equality of the triangle inequality means collinear in the same direction.
2. Compare with (A + B)².

Answer: (d) \( \vec v_1 \) and \( \vec v_2 \) are mutually perpendicular

Explanation:
1. Squaring: \( v_1^2+v_2^2+2\vec v_1\cdot\vec v_2 = v_1^2+v_2^2-2\vec v_1\cdot\vec v_2 \Rightarrow \vec v_1\cdot\vec v_2 = 0 \).

Teacher's Note:
1. Square both sides.
2. Zero dot product means perpendicular.

Answer: (b) \( -3\hat i-\hat j+5\hat k \)

Explanation:
1. Sum of the two \( = 4\hat i+\hat j-5\hat k \).
2. Required \( = \hat i-(4\hat i+\hat j-5\hat k) = -3\hat i-\hat j+5\hat k \).

Teacher's Note:
1. Subtract the current sum from the target.
2. Unit vector along x is i.

Answer: (c) \( \frac\pi2 \)

Explanation:
1. \( \vec A\times\vec B \) is perpendicular to the plane of A and B, and \( \vec A+\vec B \) lies in that plane.
2. So the angle is 90°.

Teacher's Note:
1. Cross product is normal to both vectors.
2. A + B is coplanar with A and B.

Answer: (a) \( \vec a\parallel\vec b \)

Explanation:
1. \( (ab\cos\theta)^2 = a^2b^2 \Rightarrow \cos^2\theta = 1 \), so θ = 0° or 180°: the vectors are parallel or antiparallel.

Teacher's Note:
1. cos²θ must equal 1.
2. Perpendicular vectors give zero.

Answer: (b) \( 10\hat i+10\sqrt2\hat j+10\hat k \)

Explanation:
1. Direction cosines: \( l = \cos60^\circ = \frac12 \), \( n = \cos60^\circ = \frac12 \), and \( l^2+m^2+n^2 = 1 \Rightarrow m = \frac1{\sqrt2} \).
2. \( \vec v = 20(\tfrac12,\tfrac1{\sqrt2},\tfrac12) = 10\hat i+10\sqrt2\hat j+10\hat k \).

Teacher's Note:
1. Use direction cosines.
2. Sum of their squares is 1.

Answer: (d) 64

Explanation:
1. \( 9 = 4+9+2\vec A_1\cdot\vec A_2 \Rightarrow \vec A_1\cdot\vec A_2 = -2 \).
2. Expand: \( 3A_1^2+2\vec A_1\cdot\vec A_2-8A_2^2 = 12-4-72 = -64 \), so the magnitude is 64.

Teacher's Note:
1. Square the given sum to find the dot product.
2. Expand the dot product term by term.

Answer: (c) zero, one or three (all of these are possible)

Explanation:
1. Three unit vectors at 120° to each other give zero; two opposite and one extra give 1; three in the same direction give 3.
2. So all of these are possible.

Teacher's Note:
1. Maximum is 3 when parallel.
2. Minimum is 0 when they form an equilateral triangle.

Answer: (c) 10, 10, 10 and 10, 10, 20 and 10, 20, 20 (all of these)

Explanation:
1. Resultant zero requires the largest magnitude not to exceed the sum of the other two.
2. 10 ≤ 20, 20 ≤ 20 and 20 ≤ 30 hold in all the sets, so each can be zero.

Teacher's Note:
1. Triangle inequality for vectors.
2. Equality means collinear vectors.

Answer: (d) \( 0,\ -\frac14 \) J

Explanation:
1. \( \frac{dU}{dx} = x^3-x = 0 \Rightarrow x = 0,\pm1 \).
2. \( \frac{d^2U}{dx^2} = 3x^2-1 \): at x = 0 it is −1 (local maximum, U = 0); at x = ±1 it is +2 (minimum, \( U = -\frac14 \)).

Teacher's Note:
1. Set dU/dx = 0.
2. Check the sign of the second derivative.