2 Marks Questions

Question 12 Marks

An artificial satellite is going round the earth, close to its surface. What is the time taken by it to complete one round? Given radius of the earth $=6400 km$.

Answer

Here $R =6400 km=6.4 \times 10^6 m$
$
g=9.8 ms^{-2}
$
Orbital velocity near the earth's surface is
$
v_0=\sqrt{g R}=\sqrt{9.8 \times 6.4 \times 10^6}=7918 ms^{-1}
$
Time period,
$
\begin{aligned}
& R=\frac{2 \pi R}{v 0}=\frac{2 \times 22 \times 6.4 \times 10^6}{7 \times 7918}=5080 s \\
& =1.411 \text { hour }
\end{aligned}
$

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Question 22 Marks

A cyclist speeding at $18 km / h$ on a level road takes a sharp circular turn of radius 3 m without reducing the speed. The coefficient of static friction between the tyres and the road is 0.1 . Will the cyclist slip while taking the turn?

Answer

Speed of the cyclist
$
v=18 km / h=18 \times \frac{5}{18}=5 m / s
$
Given: Radius $r=3 m \quad \mu=0.1$
The safe limit of velocity $v_s=\sqrt{\mu r g}$
So, $v_s=\sqrt{0.1 \times 10 \times 3}=1.732 m / s$
Since, cyclist rides at a faster speed than safe limit. So, the cyclist slips.

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Question 32 Marks

A famous relation in physics relates moving mass $m$ to the rest mass $m_0$ of a particle in terms of its speed $v$ and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes:
$m=\frac{m_0}{\left(1-v^2\right)^{\frac{1}{2}}}$. Guess where to put the missing c ?

Answer

The given formula will be dimensionally correct only when the dimension of L.H.S will be same as that of R.H.S. This is only possible when the factor, $\left(1-v^2\right)^{\frac{1}{2}}$ is dimensionless i.e., $\left(1-v^2\right)$ is dimensionless. This can only possible when $v^2$ is divided by the square of another velocity term. And here that term is $c^2$. So, $v ^2$ should be divided by $c^2$ to make the denominator of the fraction dimensionless. Hence, the correct relation is $m=\frac{m_0}{\left(\frac{v^2}{c^2}\right)^{\frac{1}{2}}}$ From this relativistic mass formula, one can easily see that when the velocity of a particle becomes comparable to the speed of light, its mass increases.

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Question 42 Marks

If the earth were made of lead of relative density 11.3 , what then would be the value of acceleration due to gravity on the surface of the earth? Radius of the earth $=6.4 \times 10^6 m$ and $G =6.67 \times 10^{-11} Nm ^2 kg^{-2}$.

Answer

Density of the earth,
$\rho=$ Relative density $\times$ density of water
$
=11.3 \times 10^3 kgm^{-3}
$
Acceleration due to gravity on the earth's surface,
$
\begin{aligned}
& g=\frac{G M}{R^2}=\frac{G}{R^2} \cdot \frac{4}{3} \pi R^3 \times \rho=\frac{4}{3} \pi G R \rho \\
& =\frac{4}{3} \times \frac{22}{7} \times 6.67 \times 10^{-11} \times 6.4 \times 10^6 \times 11.3 \times 10^3 \\
& =22.21 ms^{-2}
\end{aligned}
$

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Question 52 Marks

Subtract $2.5 \times 10^4$ from $3.9 \times 10^5$ with due regard to significant figures.

Answer

We have $3.9 \times 10^5-2.5 \times 10^4$
$
\begin{aligned}
& =3.9 \times 10^5-0.25 \times 10^5 \\
& =3.65 \times 10^5
\end{aligned}
$
But our answer should be rounded off up to two significant digits.
So, the answer will be $3.6 \times 10^5$

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Question 62 Marks

Write the equation of a progressive wave propagating along the positive x -direction, whose amplitude is 5 cm , frequency 250 Hz and velocity $500 ms^{-1}$.

Answer

Here $A =5 cm=0.05 m, v =250 Hz, \nu=500 ms^{-1}$
Wavelength, $\lambda=\frac{v}{\nu}=\frac{500}{250}=2 m$
Period, $T=\frac{1}{v}=\frac{1}{250} s$
The equation for the given wave can be written as $y =A \sin 2 \pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)=0.05 \sin 2 \pi\left(250 t-\frac{x}{2}\right)$
or $y=0.05 \sin \times(500 t-x)$ metre

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