Explore topic-wise InterviewSolutions in .

(c) at the extreme positions 

Explanation:

At any position tension in the string is = mg.cosθ+mv²/r (Where r is length of pendulum and v is instantaneous velocity of bob) v is zero only at extreme positions. So tension is mg.cosθ only at extreme positions.

Answer is (C) radian.

i. θ = ωt

ii. θ = \(\cfrac{2\pi t}T\)

Correct option is (C) mg tan θ 

Yes. Since acceleration is the rate of change of velocity and velocity can even be changed by changing its direction keeping the magnitude constant. So keeping the the petrol input rate constant if the motorcycle is turned along an arc of circle it will have an acceleration towards the center of arc.

Answer is (A) centripetal acceleration

Answer is (C) 12.6 m/s

Using, v = rω 

= r x (2πn) = 0.4 x 2π x 5 

= 0.4 x 2 x 3.14 x 5 

= 12.56 = 12.6 m/s

Answer is (B) 2 m/s 

Using, v = rω = 0.2 x 10 m/s = 2 m/s

The right answer is(c) N_A <N_B 

Explanation:

Both points A and B are on the crest of the curves where the weight of the car 'mg' is perpendicular to the surface. But due to the movement of the car on the curve it will feel reduction in the weight by an amount mv²/r, where 'v' is the constant speed of the car and 'r' is the radius of the curve. Clearly this reduction in weight is inversely proportional to 'r' as the numerator is constant. Smaller the 'r', greater is the term mv²/r. Since 'r' at point A is smaller than at the point B, so reduction in weight will be more at A than at B. It means the apparent weight is less at A than B. 

Normal forces at these points will be equal to the apparent weight at these points. So N_A < N_B .

The right answer is(c) N_A <N_B 

Explanation:

Both points A and B are on the crest of the curves where the weight of the car 'mg' is perpendicular to the surface. But due to the movement of the car on the curve it will feel reduction in the weight by an amount mv²/r, where 'v' is the constant speed of the car and 'r' is the radius of the curve. Clearly this reduction in weight is inversely proportional to 'r' as the numerator is constant. Smaller the 'r', greater is the term mv²/r. Since 'r' at point A is smaller than at the point B, so reduction in weight will be more at A than at B. It means the apparent weight is less at A than B. 

Normal forces at these points will be equal to the apparent weight at these points. So N_A < N_B .

The right answer is (d) 1.

Explanation: 

In completing a circle the angle covered is 2π what ever be the radius of the circle. The angular speed will be given by 2π/T where T is the time taken in covering the circle. In this case T is same for both cars, so both cars will have same angular speed ω = 2π/T . Hence ω₁/ω₂ =1

The right answer is (d) 1.

Explanation: 

In completing a circle the angle covered is 2π what ever be the radius of the circle. The angular speed will be given by 2π/T where T is the time taken in covering the circle. In this case T is same for both cars, so both cars will have same angular speed ω = 2π/T . Hence ω₁/ω₂ =1

(d) its velocity and acceleration both change.

Explanation:

Velocity and acceleration are both vectors and a vectors is fully defined with its magnitude and direction. Even when one entity changes, the vector changes. In the given case the particle moves in a circle with uniform speed. But movement in a circle makes its direction change every moment. So magnitude of the velocity is constant but its direction changes and hence the velocity changes. 

The magnitude of acceleration of the particle is given by v²/r and its direction is towards the center ie towards radius, but as the particle moves the connected radius and hence direction of the acceleration changes. So acceleration also changes as the particle moves.

Answer is  (D) v =  \(\sqrt{5lg}\)

For looping the loop, minimum velocity at the lowest point should be \(\sqrt{5lg}\)

Answer is  (D)  \(\sqrt{5lg}\)

Correct option is (B) 30

Correct Answer: (c) 30

θ = 2πn = ω_it + 1/2 at²

or 2π x 10 = 1/2 a(3)² or a = 40π / 9.

Let it make N rotations in the first 6 seconds.

2πN = 1/2 a (6)²  or N = 1/2π . 36/2 .40π/9 = 40

∴ the required number of rotations = 40 - 10 = 30.

Correct Answer is: (a, b, c, & d)

As all three particles are part of the system, any impact between them produces no change in the velocity of their centre of mass. Also, ‘A + B’ can exert a force on B only along the rod, i.e., normal to u. 

Correct option is (C) 7 m/s

Answer is (A) 2 mg

When body is released from the position (inclined at angle θ from vertical), then velocity at mean position,

v = \(\sqrt{2gl(1-cosθ)}\)

∴ Tension at the lowest point = mg + \(\frac{mv^2}{l}\) 

= mg + \(\frac{m}{l}\) [2gl(1-cos60°)] = mg + mg = 2mg

Angle of banking for a circular track = 23°2'

Factors affecting angle of banking:

i. Speed of vehicle: Angle of banking (θ) increases with maximum speed of vehicle.

ii. Radius of path: Angle of banking (θ) decreases with increase in radius of the path.

iii. Acceleration due to gravity: Angle of banking (θ) decreases with increase in the value of ‘g’.

Answer is (A) 1.88 m/s, 35.5 m/s² , 35.5 N

Linear velocity, v = ωr = 2πnr

= 2 x 3.14 x 3 x 0.1 

= 1.88 m/s

Acceleration, a = ω²r = (6π) ² x 0.1 = 35.5 m/s² 

Tension in string, T = m ω²r = 1 x (6π)² 

= 1 x (6π) ² x 0.1 

= 35.5 N

Correct option is  (C) \(2\pi\sqrt{\cfrac{l\,cos\,\theta}{g}}\)

Answer is (C) less than that of a simple pendulum of same length l.

Correct Answer is: (a) mg and T only

The force mv ²/r directed outwards, called centrifugal force, is not a real force but a pseudo force.

Correct Answer is: (b) mk²r²t 

a_c = k ²rt ² = v ²/r or v = krt.

The tangential acceleration is a_t = dv / dt = kr.

∴ the net tangential force on the particle = ma_t = mkr = F_t.

Work is done on the particle only by tangential forces, as the radial forces are perpendicular to v. 

∴ the power delivered to the particle = F_tv = (mkr)(krt) = mk 2r ²t.

The correct answer is (a) F₁ > F₂. 

Explanation:

The earth spins from west to east at equator. To maintain the same speed the first train has to press the track more than normal as it goes against the speed of earth's surface while the second train gets help from the earth's movement and it has to press the track less than the normal which explains the answer.

(d) increase at some places and remain the same at some other places.

Explanation:

If the earth stops rotating the apparent weight will increase at most of the places because the upward component of the centrifugal force due to rotation will disappear. But at poles it will not change because these places do not rotate in a circle on the earth's surface.

Correct Answer is: (a) zero

(d)  zero.

Explanation:

During the jump of the car from the cliff its trajectory is like a projectile and from the maximum height it is freely falling under gravity with an acceleration g downward. Taking it as frame of reference, it is a non-inertial frame of reference. So a pseudo force equal to mg has to be applied to the bob in upward direction which neutralizes the weight and making it zero. So tension in the string is zero.

(d)  zero.

Explanation:

During the jump of the car from the cliff its trajectory is like a projectile and from the maximum height it is freely falling under gravity with an acceleration g downward. Taking it as frame of reference, it is a non-inertial frame of reference. So a pseudo force equal to mg has to be applied to the bob in upward direction which neutralizes the weight and making it zero. So tension in the string is zero.

(a) increases

Explanation:

An over-bridge is not just an arch shape placed on a horizontal road because it will give a great jerk at the start. So to keep the driving smooth both ends of the over-bridge are kept concave upward. When a vehicle starts to ascend on this concave upward part of the over-bridge it puts extra force on the road in addition to the weight due to movement on the circular part. So the apparent weight and hence the normal force on the vehicle increases.

Answer is (D) h ≥ 5r/2

Answer is (C) internal radial acceleration.

In uniform circular motion, acceleration is caused due to change in direction and is directed radially towards centre.

Correct option is (C) 90°

Answer is (D) \(\frac{r}{2}\)

Answer is (C) 90° 

A particle will describe a circular path if the angle between velocity, \(\overrightarrow{v}\) and acceleration \(\overrightarrow{a}\) is 90°.

Answer is (A) 0

Work done by centripetal force in uniform circular motion is always equal to zero.

 (c) along a tangent 

Explanation:

When a particle moves in a circle with a speed v, its instantaneous velocity has a magnitude v and direction along the tangent at that point. The direction of the instantaneous velocity is kept changing by a centripetal force which is tension in the string in this case. As soon as the string breaks, this force disappears and the direction of the instant velocity of the particle can no longer change. So it goes along the tangent to the instantaneous point on the circle.

Answer is (A) 45°

Correct option is (D) \(\sqrt{gd}\)
 

Angle of banking = 15°13'

Elevation of the outer rail over the inner rail = 0.2625 m

Correct option is (C) a pseudo force acting along the radius and away from the centre.

Answer is (A) 10.6 km

Using, tan θ = \(\frac{v^2}{rg}\)

∴ tan 12° = \(\frac{(150)^2}{r\times10}\)

∴ r = 10.6 x 10³m = 10.6 km

Answer is (B) mv² /r

Correct option is (D) \(\cfrac{m\text v^2}r\)
 

Correct option is (A) centripetal force.

Answer is (B) mK² r² t

Centripetal acceleration

\(\frac{v^2}{r}\) = K²t²r

∴ v = K t r 

acceleration, a = \(\frac{dv}{dt}\) = \(\frac{d}{dt}\)(Ktr) = kr

F = m x a

and P = F x v = mKr x Ktr = mK² t r²