4 Marks Questions

Question 14 Marks

A person is standing on a weighing machine placed on the floor of an elevator. The elevator starts going up with some acceleration, moves with uniform velocity for a while and finally decelerates to stop. The maximum and the minimum weights recorded are $72\ kg$ and $60\ kg.$ Assuming that the magnitudes of the acceleration and the deceleration are the same, find:

The true weight of the person.
The magnitude of the acceleration. Take $g = 9.9m/s^2.$

Answer

When the elevator is accelerating upwards,
maximum weight will be recorded. $R - (W + ma ) = 0 \Rightarrow R = W + ma = m(g + a)$ max.wt.
When decelerating upwards,
maximum weight will be recorded. $R + ma - W = 0 \Rightarrow R = W - ma = m(g - a)$
So, $m(g + a) = 72 \times 9.9 …(1) m(g - a) = 60 \times 9.9 …(2)$

Now, $mg + ma = 72 \times 9.9$
$\Rightarrow mg - ma = 60 \times 9.9$
$\Rightarrow 2mg = 1306.8$
$\Rightarrow\text{m}=\frac{1306.8}{2\times9.9}=66\text{kg}$
So, the true weight of the man is $66\ kg.$
Again, to find the acceleration, $mg + ma = 72 \times 9.9$
$\Rightarrow\text{a}=\frac{72\times9.9-66\times9.9}{66}=\frac{9.9}{11}=0.9\text{m/s}^2.$

View full question & answer→

Question 24 Marks

In a $TV$ picture tube electrons are ejected from the cathode with negligible speed and reach a velocity of $5 \times 10^6m/s$ in travelling one centimeter. Assuming straight line motion, find the constant force exerted on the electron. The mass of the electron is $9.1 \times 10^{-31}kg.$

Answer

Initial velocity $u = 0 ($negligible$)$
$v = 5 \times 10^6m/s.$
$s = 1\ cm = 1 \times 10^{–2}m.$
acceleration a $=\frac{\text{v}^2-\text{u}^2}{2\text{s}}$
$=\frac{(5\times10^6)^2-0}{2\times1\times10^{-2}}$
$=\frac{25\times10^{12}}{2\times10^{-2}}$
$=12.5\times10^{14}\text{ms}^{-2}$
$F = ma $
$= 9.1 \times 10^{–31} \times 12.5 \times 10^{14} $
$= 113.75 \times 10^{–17} $
$= 1.1 \times 10^{–15}N.$

View full question & answer→

Question 34 Marks

Suppose you are running fast in a field when you suddendly find a snake in front of you. You stop quickly. Which force is responsible for your deceleration?

Answer

The force of friction acting between my feet and ground is responsible for my deceleration.

View full question & answer→

Question 44 Marks

Two blocks of unequal masses are tied by a spring. The blocks are pulled stretching the spring slightly and the system is released on a frictionless horizontal platform. Are the forces due to the spring on the two blocks equal and opposite? If yes, is it an example of Newton's third law?

Answer

Yes, the forces due to the spring on the two blocks are equal and opposite. But it’s not an example of Newton’s third law because there are three objects (2 blocks + 1 spring). Spring force on one block and force by the same block on the spring is an action-reaction pair.

View full question & answer→

Question 54 Marks

A man has fallen into a ditch of width d and two of his friends are slowly pulling him out using a light rope and two fixed pulleys as shown in figure. Show that the force (assumed equal for both the friends) exerted by each friend on the road increases as the man moves up. Find the force when the man is at a depth h.

Answer

At any depth let the ropes make angle $\theta$ with the vertical From the free body diagram

$\text{F}\cos\theta+\text{F}\cos\theta-\text{mg}=0$
$\Rightarrow2\text{F}\cos\theta=\text{mg}\Rightarrow\text{F}=\frac{\text{mg}}{2\cos\theta}$
As the man moves up. $\theta$ increases i.e. $\cos\theta$ decreases. Thus F increases.

When the man is at depth h

$\cos\theta=\frac{\text{h}}{\sqrt{\big(\frac{\text{d}}{2}\big)^2+\text{h}^2}}$
$\text{Force}=\frac{\text{mg}}{\frac{\text{h}}{\sqrt{\frac{\text{d}^2}{4}}+\text{h}^2}}=\frac{\text{mg}}{4\text{h}}\sqrt{\text{d}^2+4\text{h}^2}$

View full question & answer→

Question 64 Marks

A plumb bob is hung from the ceiling of a train compartment. If the train moves with an acceleration 'a' along a straight horizontal track, the string supporting the bob makes an angle $\tan^{-1}\Big(\frac{\text{a}}{\text{g}}\Big)$ with the normal to the ceiling. Suppose the train moves on an inclined straight track with uniform velocity. If the angle of incline is $\tan^{-1}\Big(\frac{\text{a}}{\text{g}}\Big),$ the string again makes the same angle with the normal to the ceiling. Can a person sitting inside the compartment tell by looking at the plumb line whether the train is accelerated on a horizontal straight track or it is going on an incline? If yes, how? If no, suggest a method to do so.

Answer

No, a person sitting inside the compartment can't tell jusy by looking at the plumb line whether the train is accelerating on a horizontal straight trank or moving on an incline.
When the train is accelerating along the horizontal, the tensionin the string is $\ce{mg^2 + a^2}$; when it is moving on the inclined plane, the tension is $\ce{mg}.$
So by measuring the tension in the string we can differentiate between the two cases. $\text{m}\sqrt{\text{g}^2+\text{a}^2}$

View full question & answer→

Physics 4 Marks Questions

Take a timed test