Assertion & Reason - JEE physics STD 12 Science Questions - Vidyadip

Questions

Assertion & Reason

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Question 11 Mark

Assertion : A coin is placed on phonogram turn table. The motor is started, coin moves along the moving table.
Reason : The rotating table is providing necessary centripetal force to the coin.

Answer

(d) Within a certain speed of the turn table the frictional force between the coin and the turn table supplies the necessary centripetal force required for circular motion. On further increase of speed, the frictional force cannot supply the necessary centripetal force. Therefore the coin flies off tangentially.

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Question 21 Mark

Assertion : Two similar trains are moving along the equatorial line with the same speed but in opposite direction. They will exert equal pressure on the rails.
Reason : In uniform circular motion the magnitude of acceleration remains constant but the direction continuously changes.

Answer

(e) Due to earth's axial rotation, the speed of the trains relative to earth will be different and hence the centripetal forces on them will be different. Thus their effective weights $m g-\frac{m v^2}{r}$ and $m g+\frac{m v^2}{r}$ will be different. So they exert different pressure on the rails.

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Question 31 Mark

Assertion : Cream gets separated out of milk when it is churned, it is due to gravitational force.
Reason : In circular motion gravitational force is equal to centripetal force.

Answer

(d) When the milk is churned centrifugal force acts on it outward and due to which cream in milk is separated from it.

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Question 41 Mark

Assertion : Improper banking of roads causes wear and tear of tyres.
Reason : The necessary centripetal force is provided by the force of friction between the tyres and the road.

Answer

(a) When roads are not properly banked, force of friction between tyres and road provides partially the necessary centripetal force. This cause wear and tear of tyres.

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Question 51 Mark

Assertion : A safe turn by a cyclist should neither be fast nor sharp.
Reason : The bending angle from the vertical would decrease with increase in velocity.

Answer

(c) For safe turn $\tan \theta \geq \frac{v^2}{ rg }$.It is clear that for safe turn $v$ should be small and $r$ should be large. Also bending angle from the vertical would increase with increase in velocity.

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Question 61 Mark

Assertion : When an automobile while going too fast around a curve overturns, its inner wheels leave the ground first.
Reason : For a safe turn the velocity of automobile should be less than the value of safe limit velocity.

Answer

(b) When automobile moves in circular path then reaction on inner wheel and outer wheel will be different.$R_{\text {inner }}=\frac{M}{2}\left[g-\frac{v^2 h}{r a}\right]$ and $R_{\text {outer }}=\frac{M}{2}\left[g+\frac{v^2 h}{r a}\right]$ In critical condition $v_{\text {safe }}=\sqrt{\frac{g r a}{h}}$If $v$ is equal or more that thus critical value then reaction on inner wheel becomes zero. So it leaves the ground first.

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Question 71 Mark

Assertion : If both the speed of a body and radius of its circular path are doubled, then centripetal force also gets doubled.
Reason : Centripetal force is directly proportional to both speed of a body and radius of circular path.

Answer

(c) Centripetal force is defined from formula$F=\frac{m v^2}{r} \Rightarrow F \propto \frac{v^2}{r}$If $v$ and $r$ both are doubled then $F$ also gets doubled.

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Question 81 Mark

Assertion : In circular motion, the centripetal and centrifugal force acting in opposite direction balance each other.
Reason : Centripetal and centrifugal forces don't act at the same time.

Answer

(d) While moving along a circle, the body has a constant tendency to regain its natural straight line path.This tendency gives rise to a force called centrifugal force. The centrifugal force does not act on the body in motion, the only force acting on the body in motion is centripetal force. The centrifugal force acts on the source of centripetal force to displace it radially outward from centre of the path.

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Question 91 Mark

Assertion : In circular motion, work done by centripetal force is zero.
Reason : In circular motion centripetal force is perpendicular to the displacement.

Answer

(a) We know that $W=F s \cos \theta$in the circular motion if $\theta=90^{\circ}$ then $W=0$

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Question 101 Mark

Assertion : If the speed of a body is constant, the body cannot have a path other than a circular or straight line path.
Reason : lt is not possible for a body to have a constant speed in an accelerated motion.

Answer

(d) If the speed of a body is constant, all curved paths are possible.In uniform circular motion a body has constant speed, but its direction keeps on changing, due to which it has non-zero acceleration.

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Question 111 Mark

Assertion : As the frictional force increases, the safe velocity limit for taking a turn on an unbanked road also increases.
Reason : Banking of roads will increase the value of limiting velocity.

Answer

(b) On an unbanked road, friction provides the necessary centripetal force $\frac{m v^2}{r}=\mu m g \quad \therefore v=\sqrt{\mu r g}$.Thus with increase in friction, safe velocity limit also increases.When the road is banked with angle of $\theta$ then its limiting velocity is given by $v=\sqrt{\frac{r g(\tan \theta+\mu)}{1-\mu \tan \theta}}$.\text { Thus limiting velocity increase with banking of road. }

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Question 121 Mark

Assertion : When a vehicle takes a turn on the road, it travels along a nearly circular path.
Reason : In circular motion, velocity of vehicle remains same.

Answer

(c) In circular motion the frictional force acting towards the centre of the horizontal circular path provides the centripetal force and avoid overturning of vehicle. Due to the change in direction of motion, velocity changes in circular motion.

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Question 131 Mark

Assertion : During a turn, the value of centripetal force should be less than the limiting frictional force.
Reason : The centripetal force is provided by the frictional force between the tyres and the road.

Answer

(a) The body is able to move in a circular path due to centripetal force. The centripetal force in case of vehicle is provided by frictional force. Thus if the value of frictional force $\mu mg$ is less than centripetal force, then it is not possible for a vehicle to take a turn and the body would overturn.Thus condition for safe turning of vehicle is, $ \mu m g \geq \frac{m v^2}{r}$.

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Question 141 Mark

Assertion : When a particle moves in a circle with a uniform speed, its velocity and acceleration both changes.
Reason : The centripetal acceleration in circular motion is dependent on angular velocity of the body.

Answer

(b) In uniform circular motion, the magnitude of velocity and acceleration remains same, but due to change in direction of motion, the direction of velocity and acceleration changes. Also the centripetal acceleration is given by $a=\omega^2 r$.

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Question 151 Mark

Assertion : When range of a projectile is maximum, its angle of projection may be $45^{\circ}$ or $135^{\circ}$.
Reason : Whether $\theta$ is $45^{\circ}$ or $135^{\circ}$, value of range remains the same, only the sign changes.

Answer

(a) Range, $R=\frac{u^2 \sin 2 \theta}{g}$when $\theta=45^{\circ}, R_{\max }=\frac{u^2}{g} \sin 90^{\circ}=\frac{u^2}{g}$when $\theta=135^{\circ}, R_{\max }=\frac{u^2}{g} \sin 270^{\circ}=\frac{-u^2}{g}$Negative sign shows opposite direction.

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Question 161 Mark

Assertion : The height attained by a projectile is twenty five percent of range, when projected for maximum range.
Reason : The height is independent of initial velocity of projectile.

Answer

(c) Range will be maximum when $\theta=45^{\circ}$ and in this condition $R=4 H \Rightarrow H=R / 4$ (always)because $R=4 H \cot \theta$ and $\theta=45^{\circ}$So maximum height is $25 \%$ of maximum range.It does not depends upon the velocity of projection.

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Question 171 Mark

Assertion : When the velocity of projection of a body is made $n$ times, its time of flight becomes $n$ times.
Reason : Range of projectile does not depend on the initial velocity of a body.

Answer

(c) $T \propto u$ and $R \propto u^2$When velocity of projection of a body is made $n$ times, then its time of flight becomes $n$ times and range becomes $n$ times.

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Question 181 Mark

Assertion : When a body is dropped or thrown horizontally from the same height, it would reach the ground at the same time.
Reason : Horizontal velocity has no effect on the vertical direction.

Answer

(a) Both body will take same time to reach the earth because vertical downward component of velocity for both the bodies will be zero and time of descent $t=\sqrt{\frac{2 h}{g}}$. Horizontal velocity has no effect on the vertical direction.

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Question 191 Mark

Assertion : In javelin throw, the athlete throws the projectile at an angle slightly more than $45^{\circ}$.
Reason : The maximum range does not depends upon angle of projection.

Answer

(d) If a body is projected from a place above the surface of earth, then for the maximum range, the angle of projection should be slightly less than $45^{\circ}$.

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Question 201 Mark

Assertion : The trajectory of projectile is quadratic in $y$ and linear in $x$.
Reason : $y$ component of trajectory is independent of $x$ component.

Answer

(d) $y=x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta}$

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Question 211 Mark

Assertion : For projection angle $\tan ^{-1}(4)$, the horizontal range and the maximum height of a projectile are equal.
Reason : The maximum range of projectile is directly proportional to square of velocity and inversely proportional to acceleration due to gravity.

Answer

(b) We know $R=4 H \cot \theta$
if $R=H$ then $\cot \theta=\left(\frac{1}{4}\right)$
or
$\tan \theta=(4)$and $R=\frac{u^2 \sin 2 \theta}{g}$
$\therefore R \propto \frac{u^2}{g}$

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Question 221 Mark

Assertion : Horizontal range is same for angle of projection $\theta$ and $(90-\theta)$.
Reason : Horizontal range is independent of angle of projection.

Answer

(c) Horizontal range depends upon angle of projection and it is same for complementary angles i.e. $\theta$ and $(90-\theta)$.

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Question 231 Mark

Assertion : The maximum horizontal range of projectile is proportional to square of velocity.
Reason : The maximum horizontal range of projectile is equal to maximum height attained by projectile.

Answer

(c) $R=\frac{u^2 \sin 2 \theta}{g} \therefore R_{\max }=\frac{u^2}{g}$ when $\theta=45^{\circ} \therefore R_{\max } \propto u^2$Height $H=\frac{u^2 \sin ^2 \theta}{2 g} \Rightarrow H_{\max }=\frac{u^2}{2 g}$ when $\theta=90^{\circ}$It is clear that $H_{\max }=\frac{R_{\max }}{2}$

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Question 241 Mark

Assertion : Two particles of different mass, projected with same velocity at same angles. The maximum height attained by both the particle will be same.
Reason : The maximum height of projectile is independent of particle mass.

Answer

(a) $H=\frac{u^2 \sin ^2 \theta}{2 g}$ ie. it is independent of mass of projectile.

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Question 251 Mark

Assertion : In projectile motion, the angle between the instantaneous velocity and acceleration at the highest point is $180^{\circ}\
Reason : At the highest point, velocity of projectile will be in horizontal direction only.

Answer

(e) At the highest point, vertical component of velocity becomes zero so there will be only horizontal velocity and it is perpendicular to the acceleration due to gravity.

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