2 Marks Questions

Question 12 Marks

Suppose the amplitude of a simple pendulum having a bob of mass m is $\theta_0.$ Find the tension in the string when the bob is at its extreme position.

Answer

At the extreme position, velocity of the pendulum is zero.
So there is no centrifugal force.
So, $\text{T}=\text{mg}\cos\theta_0$

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Question 22 Marks

The bob of a simple pendulum of length 1m has mass 100g and a speed of 1.4m/s at the lowest point in its path. Find the tension in the string at this instant.

Answer

At the lowest pt.
$\text{T}=\text{mg}+\frac{\text{mv}^2}{\text{r}}$
Here $\text{m}=100\text{g}=\frac{1}{10}\text{kg},\ \text{r}=1\text{m},\ \text{v}=1.4\text{m}/\text{sec}$
$\text{T}=\text{mg}+\frac{\text{mv}^2}{\text{r}}$
$\Rightarrow\frac{1}{10}\times9.8\times\frac{(1.4)^2}{10}$
$=0.98+0.196=1.176=1.2\text{N}$

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Question 32 Marks

A smooth block loosely fits in a circular tube placed on a horizontal surface. The block moves in a uniform circular motion along the tube (figure). Which wall (inner or outer) will exert a nonzero normal contact force on the block?

Answer

Outer wall will exert nonzero normal contact force on the block so to push it towards inside continuously at any instant so as to maintain its circular motion.

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Question 42 Marks

You are driving a motorcycle on a horizontal road. It is moving with a uniform velocity. Is it possible to accelerate the motoreyle without putting higher petrol input rate into the engine?

Answer

As we know F = ma in case we don’t add more petrol to engine then force applied by engine is same (constant). So to increase acceleration we can reduce mass on the bike to accelerate it more.

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Question 52 Marks

A heavy mass m is hanging from a string in equilibrium without breaking it. When this same mass is set into oscillation, the string breaks. Explain.

Answer

When the mass is at rest the tension on the string is lesser. But when is oscillates and is at its extreme position the string needs to support the mass and preventing it from going away thus the tension is much greater than rest motion.

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Question 62 Marks

A park has a radius of 10m. If a vehicle goes round it at an average speed of 18km/ hr, what should be the proper angle of banking?

Answer

Radius of Park = r = 10m Speed of vehicle = 18km/hr = 5m/ sec Angle of banking $\tan\theta=\frac{\text{v}^2}{\text{rg}}$$\Rightarrow\theta=\tan^{-1}\frac{\text{v}^2}{\text{rg}}$
$\theta=\tan^{-1}\frac{25}{100}=\tan^{-1}\Big(\frac{1}{4}\Big)$

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Question 72 Marks

A circular road of radius 50m has the angle of banking equal to 30°. At what speed should a vehicle go on this road so that the friction is not used?

Answer

Angle of banking $=\theta=30^\circ$ Radius $=\text{r}=50\text{m}$$\tan\theta=\frac{\text{v}^2}{\text{rg}}$
$\Rightarrow\tan30^\circ=\frac{\text{v}^2}{\text{rg}}$
$\Rightarrow\frac{1}{\sqrt3}=\frac{\text{v}^2}{\text{rg}}$
$\Rightarrow\text{v}^2=\frac{\text{rg}}{\sqrt3}=\frac{50\times10}{\sqrt3}$
$\Rightarrow\text{v}=\sqrt{\frac{500}{\sqrt3}}=17\text{m}/\text{sec}.$

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Question 82 Marks

If the road of the previous problem is horizontal (no banking),The road is horizontal (no banking) what should be the minimum friction coefficient so that a scooter going at 18km/hr does not skid?

Answer

The road is horizontal (no banking)
$\frac{\text{mv}^2}{\text{R}}=\mu\text{N}$
and N = mg
So $\frac{\text{mv}^2}{\text{R}}=\mu\text{mg,}\ \text{v}=5\text{m}/\text{sec},\text{R}=10\text{m}$
$\Rightarrow\frac{25}{10}=\mu\text{g}$
$\Rightarrow\mu=\frac{25}{100}=0.25$

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