2 Marks Questions

Question 12 Marks

The acceleration due to gravity on the surface of the earth is $10 ms^{-2}$. The mass of the planet Mars as compared to earth is $\frac{1}{10}$ and radius is $\frac{1}{2}$. Determine the gravitational acceleration of a body on the surface of Mars.

Answer

$\begin{aligned} & g _{ e }=\frac{G M_e}{r_e^2} \\ & g_{ m }=\frac{G M_m}{r_m^2} \\ & \frac{g_e}{g_m}=\frac{M_e}{M_m} \times \frac{r_m^2}{r_e^2}\end{aligned}$
$\begin{aligned} & \frac{g_e}{g_m}=\frac{10}{4} \\ & g_m=4 ms^{-2}\end{aligned}$

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Question 22 Marks

Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the midpoint of the line joining the centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?

Answer

Gravitational potential at the midpoint of the line joining the centres of the two spheres is
$
\begin{aligned}
& V=\frac{G M}{\frac{T}{2}}-\frac{G M}{\frac{\tau}{2}}=\frac{4 G M}{r} \\
& V=\frac{4 \times 6.67 \times 10^{-11} \times 100}{0.1} \\
& =-2.68 \times 10^{-7} J / kg
\end{aligned}
$
As the effective force on the body placed at mid-point is zero, so the body is in equilibrium. If the body is displaced a little towards either mass body from its equilibrium position, it will not return back to its initial position of equilibrium. Hence, the body is in an unstable equilibrium.

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Question 32 Marks

A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei, the products must be emitted in opposite directions.

Answer

According to the principle of conservation of linear momentum, total momentum remains constant.
Before disintegration linear momentum $=$ zero
After disintegration linear momentum $=m_1 \overline{v_1}+m_2 \overline{v_2}$
$
\Rightarrow m_1 v_1+m_2 v_2=0 \Rightarrow v_2=-\frac{m_1 v_1}{m_2}
$

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Question 42 Marks

A small steel ball of radius $r$ is allowed to fall under gravity through a column of a viscous liquid of coefficient of viscosity $\eta$. After some time the velocity of the body attains a constant value $v _{ T }$. The terminal velocity depends upon (i) the weight of the ball mg (ii) the coefficient of viscosity $\eta$ and (iii) the radius of the ball r. By the method of dimensions, determine the relation expressing terminal velocity.

Answer

Let $v _{ T }=K(m g)^a \eta^b r^c$
where $K = a$ dimensionless constant.
Putting the dimensions of various quantities,
$LT ^{-1}=\left[ MLT ^{-2}\right]^{ a }\left[ ML ^{-1} T^{-1}\right]^{ b }[ L ]^{ c }$
or $M^0 L^1 T^{-1}=M^{a+b} L^{a-b+c} T^{-2 a-b}$
Equating the powers of $M , L$ and T on both sides, we get
$
a+b=0, a-b+c=1,-2 a-b=-1
$
On solving, $a=1, b=-1, c=-1$
$
\therefore v_{T}=K(mg)^1 \eta^{-1} r^{-1} \text { or } v_T \propto \frac{m g}{\eta r}
$

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Question 52 Marks

If the value of universal gravitational constant in SI is $6.6 \times 10^{-11} Nm ^2 kg^{-2}$, then find its value in CGS system.

Answer

$
\begin{aligned}
& G=6.6 \times 10^{-11} Nm^2 kg^{-2} \\
& =6.6 \times 10^{-11} Ng^{-1} m^3 kg^{-2}
\end{aligned}
$
since,
$
\begin{aligned}
& 1 kg=10^3 g \\
& 1 m=10^2 cm
\end{aligned}
$
Hence
$
\begin{aligned}
& =6.6 \times 10^{-11}\left(10^3 g\right)\left(10^2 cm\right) 3 s^{-2} \\
& =6.6 \times 10^{-11-3+3} g^{-1} cm^3 s^{-2} \\
& =6.6 \times 10^{-8} g^{-1} cm^3 s^{-2}
\end{aligned}
$

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Question 62 Marks

State few important uses of the phenomenon of beats.

Answer

Some important uses of beats phenomenon are as follows:
i. Principle of beats enables us to tune one musical instrument by sounding it against a standard frequency.
ii. We may determine the frequency of a tuning fork by studying beats formed with another tuning fork of known frequency.
iii. Principle of beats is made use of in heterodyne method of radio reception.

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