JEE Physics Basic Mathematics Used in Physics MCQs Set 04

Answer: (a) \( 2 \)

Explanation:
1. This is an infinite geometric series with first term \( a=1 \) and common ratio \( r=\frac12 \).
2. Since \( |r|<1 \), \( S_\infty=\frac{a}{1-r} \).
3. \( S_\infty=\frac{1}{1-\frac12}=2 \).

Teacher's Note:
1. An infinite G.P. has a finite sum only when the common ratio satisfies |r| < 1.
2. The same series appears in physics when summing successive bounces or reflections.

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

Explanation:
1. Substitute \( u=2x-3 \), so \( du=2\,dx \).
2. \( \int\frac{dx}{2x-3}=\frac12\int\frac{du}{u}=\frac12\log_e u+c \).
3. Hence the result is \( \frac{\log_e(2x-3)}{2}+c \).

Teacher's Note:
1. The integral of 1/(ax + b) is (1/a) times the natural log of (ax + b).
2. Do not forget the factor 1/a that comes from the chain rule.

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

Explanation:
1. For \( x\ll1 \), the binomial approximation gives \( (1+x)^n\approx1+nx \).
2. Here \( x=\frac hR \) and \( n=-2 \).
3. So \( \left(1+\frac hR\right)^{-2}\approx1-\frac{2h}{R} \).

Teacher's Note:
1. Neglect squares and higher powers of the small quantity h/R.
2. This is exactly how g at a small height h is written as g(1 - 2h/R).

Answer: (a) \( \frac32,\ \frac52 \)

Explanation:
1. For \( ax^2+bx+c=0 \), sum of roots \( =-\frac ba \) and product \( =\frac ca \).
2. Here \( a=2,\ b=-3,\ c=5 \).
3. Sum \( =\frac32 \) and product \( =\frac52 \).

Teacher's Note:
1. Sum and product of roots can be written without actually solving the equation.
2. Do not forget the negative sign in the formula for the sum.

Answer: (d) 0.999

Explanation:
1. Write \( 0.996=1-0.004 \).
2. By the binomial approximation, \( (1-x)^{1/4}\approx1-\frac x4 \) with \( x=0.004 \).
3. \( (0.996)^{1/4}\approx1-0.001=0.999 \).

Teacher's Note:
1. The approximation works because 0.004 is much smaller than 1.
2. Check the answer by raising 0.999 to the power 4: the result is about 0.996.

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

Explanation:
1. Let \( u=4x \), so \( dx=\frac{du}{4} \).
2. \( \int\sin u\,\frac{du}{4}=-\frac{\cos u}{4}+c \).
3. Result: \( -\frac{\cos4x}{4}+c \).

Teacher's Note:
1. When the argument is ax, the integral picks up a factor 1/a.
2. Differentiate the answer to check that you get sin 4x back.

Answer: (a) -5 unit

Explanation:
1. \( \frac{d\phi}{dt}=4t-3 \).
2. At \( t=2 \): \( \frac{d\phi}{dt}=4(2)-3=5 \).
3. \( e=-\frac{d\phi}{dt}=-5 \) unit.

Teacher's Note:
1. Induced emf is the negative rate of change of flux (Faraday and Lenz).
2. Differentiate first, then substitute the value of t.

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

Explanation:
1. \( 480^\circ=360^\circ+120^\circ \), so \( \sin480^\circ=\sin120^\circ \).
2. \( \sin120^\circ=\sin(180^\circ-60^\circ)=\sin60^\circ \).
3. Hence \( \sin480^\circ=\frac{\sqrt3}{2} \).

Teacher's Note:
1. Subtract full rotations of 360 degrees first.
2. In the second quadrant sine is positive, cosine is negative.

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

Explanation:
1. Use the product rule \( \frac{d}{dx}(uv)=u'v+uv' \) with \( u=x,\ v=\sin x \).
2. \( \frac{dy}{dx}=1\cdot\sin x+x\cos x \).

Teacher's Note:
1. The product rule must be used whenever two functions of x are multiplied.
2. Many physics formulas, such as power P = Fv, need this rule.

Answer: (a) \( \frac\theta2+\frac{\sin2\theta}{4}+c \)

Explanation:
1. Use \( \cos^2\theta=\frac{1+\cos2\theta}{2} \).
2. \( \int\frac{1+\cos2\theta}{2}d\theta=\frac\theta2+\frac{\sin2\theta}{4}+c \).

Teacher's Note:
1. Reduce powers of sine and cosine with the double-angle identity before integrating.
2. The average of cos squared over a full cycle is 1/2, which comes from the first term.

Answer: (d) \( 4\cos(4x-3) \)

Explanation:
1. Use the chain rule: differentiate the outer function sine, then multiply by the derivative of the inner function \( 4x-3 \).
2. \( \frac{dy}{dx}=\cos(4x-3)\times4=4\cos(4x-3) \).

Teacher's Note:
1. The chain rule multiplies by the derivative of the inside function.
2. Displacement x = A sin(wt + phi) gives velocity A w cos(wt + phi) by the same rule.

Answer: (a) slope \( \frac32 \), intercept +2

Explanation:
1. Rewrite in the form \( y=mx+c \): \( y=\frac32x+2 \).
2. Slope \( m=\frac32 \) (positive, the line rises to the right) and y-intercept \( c=+2 \).

Teacher's Note:
1. Always convert the equation to y = mx + c to read off the slope and intercept.
2. A positive slope with a positive intercept means the line cuts the y-axis above the origin and rises.

Answer: (b) \( y\propto x^2 \)

Explanation:
1. A parabola opening along the y-axis with vertex at the origin has the equation \( y=kx^2 \).
2. Hence \( y\propto x^2 \).

Teacher's Note:
1. If the parabola opened along the x-axis the relation would be y squared proportional to x.
2. Compare with free fall: distance is proportional to the square of time.

Answer: (a) a straight line with positive slope, cutting the a-axis below the origin (negative intercept)

Explanation:
1. \( a=\frac{dv}{dt}=8t-2 \).
2. This is linear in t, with slope \( +8 \) and a-intercept \( -2 \).
3. So the graph is a straight line with positive slope that cuts the a-axis below the origin.

Teacher's Note:
1. Differentiating a quadratic always gives a straight line.
2. The intercept is the value of the acceleration at t = 0.

Answer: (b) a parabola opening downward, passing through the origin and cutting the x-axis again at x = 1/2

Explanation:
1. The coefficient of \( x^2 \) is negative, so the parabola opens downward.
2. Putting \( x=0 \) gives \( y=0 \): it passes through the origin.
3. Putting \( y=0 \) gives \( 2x(1-2x)=0 \), so \( x=0 \) or \( x=\frac12 \).
4. Maximum occurs at \( x=\frac14 \) with \( y=\frac14 \).

Teacher's Note:
1. The sign of the x squared term decides whether the parabola opens up or down.
2. Find the x-intercepts by factorising, and the vertex from dy/dx = 0.

Answer: (c) 4 m/s

Explanation:
1. Average value \( =\frac{\int_0^1(2t+3)dt}{1-0} \).
2. \( \int_0^1(2t+3)dt=\left[t^2+3t\right]_0^1=4 \).
3. Average velocity \( =4 \) m/s.

Teacher's Note:
1. For a linear function, the average is also the mean of the end values: (3 + 5)/2 = 4.
2. Use the integral for the mean when the function is not linear.

Answer: (b) \( -3\le x\le3 \)

Explanation:
1. Kinetic energy cannot be negative, so \( 9-x^2\ge0 \).
2. \( x^2\le9 \), that is \( -3\le x\le3 \).

Teacher's Note:
1. A physical restriction such as K at least 0 translates into a mathematical inequality.
2. The particle turns back at the points where K = 0, that is x = plus or minus 3.

Answer: (b) \( 5,\ \sin^{-1}\left(\frac35\right) \)

Explanation:
1. Expand: \( A\sin(\theta+\alpha)=A\cos\alpha\sin\theta+A\sin\alpha\cos\theta \).
2. Compare: \( A\cos\alpha=4 \) and \( A\sin\alpha=3 \).
3. Squaring and adding gives \( A=5 \), and \( \sin\alpha=\frac35 \).

Teacher's Note:
1. The expression a cos t + b sin t always has amplitude equal to the root of (a squared + b squared).
2. Here the 3-4-5 triangle fixes the angle.

Answer: (b) \( 2\ell\sin\frac\theta2 \)

Explanation:
1. The initial and final vectors have equal magnitude \( \ell \) with angle \( \theta \) between them.
2. Change \( =|\vec A_2-\vec A_1|=\sqrt{\ell^2+\ell^2-2\ell^2\cos\theta} \).
3. \( =\ell\sqrt{2(1-\cos\theta)}=2\ell\sin\frac\theta2 \).

Teacher's Note:
1. The change is the third side of an isosceles triangle with two sides of length l.
2. For a small angle this reduces to the arc length l times theta.

Answer: (a) 6 N and 10 N

Explanation:
1. Let the smaller force be \( a \) and the other \( b \), with \( a+b=16 \).
2. The resultant \( R=8 \) is perpendicular to \( a \), so \( b^2=a^2+R^2 \), giving \( b^2-a^2=64 \).
3. \( (b-a)(b+a)=64 \Rightarrow b-a=4 \).
4. So \( b=10 \) N and \( a=6 \) N.

Teacher's Note:
1. When the resultant is perpendicular to one force, the three vectors form a right-angled triangle.
2. Check: 6, 8 and 10 form a Pythagorean triplet.

Answer: (b) \( \theta_1=\frac{\theta_2}{2} \)

Explanation:
1. First case: \( P,Q,R \) equal in magnitude and \( \vec P+\vec Q=\vec R \). The three vectors form an equilateral triangle with \( \vec P \) and \( \vec Q \) at \( 120^\circ \); \( \vec R \) bisects them, so \( \theta_1=60^\circ \).
2. Second case: three equal vectors with zero sum are at \( 120^\circ \) to each other, so \( \theta_2=120^\circ \).
3. Hence \( \theta_1=\frac{\theta_2}{2} \).

Teacher's Note:
1. Three equal vectors that sum to zero are always 120 degrees apart.
2. Draw the vector triangle in each case to see the angle.

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

Explanation:
1. The first ball moves at \( \tan^{-1}\frac{\sqrt3}{1}=60^\circ \) to the x-axis.
2. The second ball moves at \( \tan^{-1}\frac22=45^\circ \) to the x-axis.
3. Angle between the paths \( =60^\circ-45^\circ=15^\circ \).

Teacher's Note:
1. The direction of a velocity vector is found from the ratio of its components.
2. The angle between two paths is the difference of their angles with a common axis.

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

Explanation:
1. The tip moves in a circle of radius 1 cm with period 60 s, so \( \omega=\frac{2\pi}{60}=\frac{\pi}{30} \) rad/s and speed \( v=\omega r=\frac{\pi}{30} \) cm/s.
2. In 15 s the hand turns through a quarter circle (\( 90^\circ \)), so the velocity vector turns through \( 90^\circ \).
3. \( |\Delta\vec v|=\sqrt{v^2+v^2}=\sqrt2\,v=\frac{\pi}{30}\sqrt2 \) cm/s.

Teacher's Note:
1. The speed is constant but the velocity changes because its direction changes.
2. For a turn through theta, the change in velocity is 2 v sin(theta/2).