2 Marks Questions

Question 12 Marks

If the mass of the sun is $2 \times 10^{30} kg$, the distance of the earth from the sun is $1.5 \times 10^{11} m$ and period of revolution of the earth around the sun is one year ( $=365.3$ days), calculate the value of gravitational constant.

Answer

$
\begin{aligned}
& G \text { (gravitational constant) }=6.673 \times 10^{-11} Nm^2 kg^{-2} \\
& M(\text { mass of sun })=2 \times 10^{30} kg \\
& m \text { (mass of earth) }=6 \times 10^{24} kg
\end{aligned}
$
d (average distance between sun and earth) $=1.5 \times 10^8 km=1.5 \times 10^8 \times 1000=1.5 \times 10^{11} m$
$
\begin{aligned}
& F=\frac{G M m}{d^2} \\
& =\frac{6.67 \times 10^{-11} \times 2 \times 10^{30} \times 6 \times 10^{24}}{\left(1.5 \times 10^{11}\right)^2} \\
& =\frac{6.673 \times 10^{-11} \times 2 \times 10^{30} \times 6 \times 10^{24}}{1.5 \times 1.510^{22}}\left(10^{24+30-11}=10^{43}\right) \\
& =\frac{6.673 \times 2 \times 6 \times 10^{43}}{1.5 \times 1.5 \times 10^{22}}\left(10^{43-22}=10^{21}\right) \\
& =\frac{6.673 \times 2 \times 6 \times 10^{21}}{1.5 \times 1.5}\left(10^{43-22}=10^{21}\right) \\
& =\frac{80.076 \times 10^{21}}{1.5 \times 1.5} \\
& =35.59 \times 10^{21} N=3.57 \times 10^{22} N
\end{aligned}
$
Force of gravity $=3.57 \times 10^{22} N$

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Question 22 Marks

Define the term wave motion. Give four important characteristics of wave motion.

Answer

Wave motion is a form of disturbance which travels through a medium due to repeated periodic motion of the particles of the medium about their mean positions.
Four important characteristics of wave motion are:
i. Wave motion is a form of disturbance which travels through a medium due to the vibrations of the particles of the medium.
ii. It is the disturbance which travels in the forward direction and not the particles. The particles simply vibrate about their mean position.
iii. The energy of a particle is wholly kinetic at the mean position and wholly potential at the extreme position.
iv. The motion of each particle begins a little later than that of its predecessor. In other words, there is always a constant phase difference between any two neighbouring particles. The wave always advances in that direction in which it meets particles with decreasing phase.

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Question 32 Marks

Calculate the energy required to move an earth satellite of mass $10^3 kg$ from a circular orbit of radius 2 R to that of radius 3R. Given mass of the earth, $M=5.98 \times 10^{24} kg$ and radius of the earth, $R=6.37 \times 10^6 m$.

Answer

Total energy of a satellite,
$
\begin{aligned}
& E=\text { P.E. }+ \text { K.E. }=-\frac{G M m}{r}+\frac{1}{2} m v^2 \\
& =-\frac{GMm}{r}+\frac{1}{2} m \cdot \frac{GM}{r}=-\frac{GMm}{2 r} \\
& W=E_{f}-E_{i}=-\frac{GMm}{2 \times 3 R}+\frac{GMm}{2 \times 2 R}=\frac{GMm}{12 R} \\
& =\frac{6.67 \times 10^{11} \times 5.98 \times 10^{24} \times 10^3}{12 \times 6.37 \times 10^6} \\
& =5.02 \times 10^9 J
\end{aligned}
$

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Question 42 Marks

A rocket with a lift-off mass $20,000 kg$ is blasted upwards with an initial acceleration of $5.0 ms^{-2}$. Calculate the initial thrust (force) of the blast.

Answer

Here, lift-off mass $m =20,000 kg$ and $a =5.0 m s ^{-2}$ vertically upward.
Using Newton's second of motion, the net force (thrust) acting on the rocket is given by the relation:
$
\begin{aligned}
& F-m g=m a \\
& F=m g+m a=m(g+a) \\
& =20,000(10+5) \\
& =20,000 \times 15 \\
& =3 \times 10^5 N
\end{aligned}
$

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Question 52 Marks

Calculate the dimensions of force and impulse taking velocity, density and frequency as basic quantities.

Answer

$
v=LT^{-1}, \rho=ML^{-3}, v=T^{-1}
$
Solving for $M , L$ and T in terms of $v , \rho$ and $v$, we get
$
\begin{aligned}
& T=v^{-1}, L=v \nu^{-1} l, M=\rho v^3 \nu^{-3} \\
& \text { [Force] }=MLT^{-2}=\rho v^3 \nu^{-3} v \nu^{-1} v^2=\rho v^4 \nu^{-2} \\
& \text { [Impulse] = Force } \times \text { time }=\rho v^4 \nu^{-2} v^{-1}=\rho v^4 \nu^{-3}
\end{aligned}
$

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Question 62 Marks

Assuming that the critical velocity $v _{ C }$ of a viscous liquid flowing through a capillary tube depends only upon the radius $r$ of the tube, density $\rho$ and the coefficient of viscosity $\eta$ of the liquid, find the expression for critical velocity.

Answer

Velocity of viscous liquid is flowing through a capillary tube depends on radius r of tube, density $\rho$, and coefficient of viscosity of the liquid.
Let the critical velocity is proportional to $r ^{ a }, \rho^b, \eta^n$.
Then,
$V = kr ^{ a }, \rho^b, \eta^n, k$ is any dimension less constant.
Thus, Dimension of $V = LT ^{-1}$
Dimension of $r = L$
Dimension of $\rho= ML ^{-3}$
Dimension of $\eta= ML ^{-1} T^{-1}$
Now, Putting dimension on both sides of equation,
$
\begin{aligned}
& {\left[LT^{-1}\right]=[L]^{a}\left[ML^{-3}\right]^{b}\left[ML^{-1} T^{-1}\right]^{n}} \\
& {\left[M^0 LT^{-1}\right]=\left[M^{b+n}\right]\left[L^{a-n-3 b}\right]\left[T^{-n}\right]}
\end{aligned}
$
On comparing,
$
\begin{aligned}
& n=1, a-n-3 b=1 \text { and } b+n=0 \\
& b=-n=-1
\end{aligned}
$
Now, in a - $n -3 b=1$,
$
\begin{aligned}
& a-1-3(-1)=1 \\
& a+2=1 \\
& a=-1 \\
& V=\frac{k \eta}{r^2 \rho}
\end{aligned}
$

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