PhysicsQ1–3 of 3 questions
Page 1 of 1 · English
1
PhysicsEasyMCQIIT JEE · 1973
$A$ lorry and a car moving with the same kinetic energy $(K.E.)$ are brought to rest by applying the same retarding force. Then:
A
Lorry will come to rest in a shorter distance
B
Car will come to rest in a shorter distance
C
Both come to rest in the same distance
D
None of the above
Solution
(C) According to the work-energy theorem,the work done by the retarding force $(F)$ is equal to the change in kinetic energy of the vehicle.
$W = \Delta K.E.$
Since the vehicles are brought to rest,the work done $W = F \times s$,where $s$ is the stopping distance.
Therefore,$F \times s = K.E.$
$s = \frac{K.E.}{F}$
Since both the lorry and the car have the same initial kinetic energy $(K.E.)$ and are subjected to the same retarding force $(F)$,the stopping distance $(s)$ must be the same for both vehicles.
$W = \Delta K.E.$
Since the vehicles are brought to rest,the work done $W = F \times s$,where $s$ is the stopping distance.
Therefore,$F \times s = K.E.$
$s = \frac{K.E.}{F}$
Since both the lorry and the car have the same initial kinetic energy $(K.E.)$ and are subjected to the same retarding force $(F)$,the stopping distance $(s)$ must be the same for both vehicles.
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2
PhysicsEasyMCQIIT JEE · 1973
$A$ particle executes simple harmonic motion with a frequency $f$. The frequency with which its kinetic energy oscillates is
A
$f/2$
B
$f$
C
$2f$
D
$4f$
Solution
(C) The displacement of a particle in simple harmonic motion is given by $x = a \sin(\omega t)$.
The velocity is $v = \frac{dx}{dt} = a\omega \cos(\omega t)$.
The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2 = \frac{1}{2}m(a\omega \cos(\omega t))^2$.
Using the trigonometric identity $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$,we get:
$K = \frac{1}{2}m a^2 \omega^2 \cos^2(\omega t) = \frac{1}{4}m a^2 \omega^2 (1 + \cos(2\omega t))$.
The term $\cos(2\omega t)$ indicates that the kinetic energy oscillates with an angular frequency of $2\omega$.
Since $\omega = 2\pi f$,the frequency of oscillation of kinetic energy is $2f$.
The velocity is $v = \frac{dx}{dt} = a\omega \cos(\omega t)$.
The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2 = \frac{1}{2}m(a\omega \cos(\omega t))^2$.
Using the trigonometric identity $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$,we get:
$K = \frac{1}{2}m a^2 \omega^2 \cos^2(\omega t) = \frac{1}{4}m a^2 \omega^2 (1 + \cos(2\omega t))$.
The term $\cos(2\omega t)$ indicates that the kinetic energy oscillates with an angular frequency of $2\omega$.
Since $\omega = 2\pi f$,the frequency of oscillation of kinetic energy is $2f$.
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3
PhysicsMediumMCQIIT JEE · 1973
The mass and diameter of a planet are twice those of earth. The period of oscillation of a pendulum on this planet will be (if it is a second's pendulum on earth).
A
$\frac{1}{\sqrt{2}} \, \text{s}$
B
$2\sqrt{2} \, \text{s}$
C
$2 \, \text{s}$
D
$\frac{1}{2} \, \text{s}$
Solution
(B) The acceleration due to gravity is given by $g = \frac{GM}{R^2}$.
Given: $M_p = 2M_e$ and $R_p = 2R_e$.
Therefore,the ratio of gravity on earth to the planet is:
$\frac{g_e}{g_p} = \frac{M_e}{M_p} \times \left(\frac{R_p}{R_e}\right)^2 = \frac{1}{2} \times (2)^2 = \frac{4}{2} = 2$.
So,$g_p = \frac{g_e}{2}$.
The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$,which implies $T \propto \frac{1}{\sqrt{g}}$.
For a second's pendulum on earth,$T_e = 2 \, \text{s}$.
Thus,$\frac{T_p}{T_e} = \sqrt{\frac{g_e}{g_p}} = \sqrt{2}$.
$T_p = T_e \times \sqrt{2} = 2\sqrt{2} \, \text{s}$.
Given: $M_p = 2M_e$ and $R_p = 2R_e$.
Therefore,the ratio of gravity on earth to the planet is:
$\frac{g_e}{g_p} = \frac{M_e}{M_p} \times \left(\frac{R_p}{R_e}\right)^2 = \frac{1}{2} \times (2)^2 = \frac{4}{2} = 2$.
So,$g_p = \frac{g_e}{2}$.
The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$,which implies $T \propto \frac{1}{\sqrt{g}}$.
For a second's pendulum on earth,$T_e = 2 \, \text{s}$.
Thus,$\frac{T_p}{T_e} = \sqrt{\frac{g_e}{g_p}} = \sqrt{2}$.
$T_p = T_e \times \sqrt{2} = 2\sqrt{2} \, \text{s}$.
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