2 Marks Questions

Question 12 Marks

Assuming the earth to be a uniform sphere of radius 6400 km and density $5.5 gcm ^{-3}$, find the value of g on its surface. Given G $=6.66 \times 10^{-11} Nm ^2 kg^{-2}$.

Answer

$\begin{aligned} & \text { Given } \rho=5.5 g / cc \\ & =\frac{5.5 \times 10^{-3} Kg }{\left(10^{-2} m\right)^3}=5.5 \times 10^3 kg / m ^3 \\ & R =6400 Km =6.4 \times 10^6 m \\ & \therefore G =\frac{3 \times 9.8}{4 \times 3.14 \times\left(6.4 \times 10^6\right) \times 5.5 \times 10^3} \\ & =6.6 \times 10^{-11} \frac{ Nm ^2}{kg^2}\end{aligned}$

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Question 22 Marks

Derive an expression for work done against gravity.

Answer

Potential energy of the body on the surface of the earth $=\frac{-G M m}{R}$
Potential energy at a height $h$ from the surface of the earth $=-\frac{G M m}{(R+h)}$
$
\begin{aligned}
& \text { Work done }=\left(-\frac{G M m}{R+h}\right)-\left(-\frac{G M m}{R}\right) \\
& =\frac{G M m}{R}-\frac{G M m}{R+h} \\
& =G M m\left(\frac{1}{R}-\frac{1}{R+h}\right) \\
& =\frac{G M m h}{R(R+h)}=\frac{M g R^2 h}{R(R+h)}\left[\because g=\frac{G M}{R^2}\right] \\
& =\frac{(M g h) R}{(R+h)}=\frac{M g h}{1+\frac{h}{R}}
\end{aligned}
$

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Question 32 Marks

In the Atwood's machine (figure), the system starts from rest. What is the speed and distance moved by each mass at t = 3s?

Answer

The speed will be same for both block
so let us consider the block 1
Here the force on the block is
$
F=\left(m_1-m_2\right) g=(12-10) \times 9.8=19.6 N
$
So the acceleration of the system is a $=\frac{F}{m_1-m_2}=0.6125$
So the speed after 3 s will be $0.6125 \times 3=1.84 ms^{-1}$

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Question 42 Marks

If force F, length L and time T are taken as fundamental units then what will be the dimensions of mass?

Answer

Let $m=K F^a L^b T^c$
Substituting the dimension of, $[F]=\left[M L T^{-2}\right],[L]=[L]$ and $[T]=[T]$, we have
$
\begin{aligned}
& {[M]=\left[MLT^{-2}\right]^{a}[L]^{b}[T]^{c}} \\
& {[M]=M^{a} L^{a+b} T^{-2 a+c}}
\end{aligned}
$
On equating the powers on both sides, we get
$
a=1, a+b=0,-2 a+c=0
$
Solvign these equations, we get
$
a=1, b=-1 \text { and } c=2
$
Hence, dimensions of mass M are $\left[ F ^1 L^{-1} T^2\right]$.

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Question 52 Marks

Convert:
i. $3.0 m / s ^2=$ __________ $km / hr ^2$
ii. $6.6710^{-11} Nm ^2 / kg ^2=$ __________ $g ^{-1} cm^3 s^{-2}$

Answer

$\begin{aligned} & \text { i. } 1 \text { hour }=3600 sec \text { so that } 1 sec =1 / 3600 \text { hour } \\ & 1 km=1000 m \text { so that } 1 m=1 / 1000 km \\ & 3.0 m s ^{-2}=3.0(1 / 1000 km)(1 / 3600 \text { hour })^{-2} \\ & =3.9 \times 10^4 km / hr ^2 \\ & \text { ii. } 6.6710^{-11} Nm ^2 / kg ^2= g ^{-1} cm^3 s^{-2} \\ & =6.67 \times 10^{-11} kg^{-1} m^3 s^{-2} \\ & =6.67 \times 10^{-11} \times 10^3 \times\left(10^2\right)^3 \\ & =6.67 \times 10^{-8} g^{-1} cm^3 s^{-2}\end{aligned}$

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Question 62 Marks

A steel wire has a length of 12.0 m and a mass of 2.10 kg . What should be the tension in the wire so that the speed of a transverse wave on the wire equals the speed of sound in dry air at $20^{\circ} C =343 ms^{-1}$ ?

Answer

Length of the steel wire, $l =12 m$
Mass of the steel wire, $m =2.10 kg$
Velocity of the transverse wave, $v =343 m / s$
Mass per unit length, $\mu=\frac{m}{l}=\frac{2.10}{12}=0.175 kgm ^{-1}$
For tension $T$, the velocity of the transverse wave can be obtained using the relation:
$v=\sqrt{\frac{T}{\mu}}$ here T is tension in the wire
$
\begin{aligned}
& \therefore T=v^2 \mu \\
& =(343)^2 \times 0.175=20588.575 \approx 206 \times 10^4 N
\end{aligned}
$

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