3 Marks Question

Question 13 Marks

Find the acceleration of the moon with respect to the earth from the following data:
Distance between the earth and the moon $= 3.85 \times 10^5\ km$ and the time taken by the moon to complete one revolution around the earth $= 27.3$ days.

Answer

Distance between Earth $\&$ Moon,$\text{r}=3.85\times10^5\text{km}=3.85\times10^8\text{m}$
$\text{T}=27.3\ \text{days}=24\times3600\times(27.3)\text{sec}$
$=2.36\times10^6\text{sec}$
$\text{v}=\frac{2\pi\text{r}}{\text{T}}=\frac{2\times3.14\times3.85\times10^8}{2.36\times10^6}$
$=1025.42\text{m}/\text{sec}$
$\text{a}=\frac{\text{v}^2}{\text{r}}=\frac{(1025.42)^2}{3.85\times10^8}$
$=0.00273\ \text{m}/\text{sec}^2$
$=2.73\times10^{-3}\ \text{m}/\text{sec}^2$

View full question & answer→

Question 23 Marks

If the horizontal force needed for the turn in the previous problem is to be supplied by the normal force by the road, what should be the proper angle of banking?

Answer

in the diagram

$\text{R}\cos\theta=\text{mg}\ \dots(1)$
$\text{R}\sin\theta=\frac{\text{mv}^2}{\text{r}}\ \dots(2)$
Dividing equation (1) with equation (2),$\tan\theta=\frac{\text{mv}^2}{\text{rmg}}=\frac{\text{v}^2}{\text{rg}}$
$\text{v}=36\text{km}/\text{hr}=10\text{m}/\text{sec},\text{r}=30\text{m}$
$\tan\theta=\frac{\text{v}^2}{\text{rg}}$
$=\frac{100}{30\times10}=\frac{1}{3}$
$\Rightarrow\theta=\tan^{-1}\Big(\frac{1}{3}\Big)$

View full question & answer→

Question 33 Marks

A ceiling fan has a diameter (of the circle through the outer edges of the three blades) of 120cm and rpm 1500 at full speed. Consider a particle of mass 1g sticking at the outer end of a blade. How much force does it experience when the fan runs at full speed? Who exerts this force on the particle? How much force does the particle exert on the blade along its surface?

Answer

A celling fan has a diameter = 120cm.$\therefore$ Radius = r = 60cm = 0.6
Mass of particle on the outer end of a blade is 1g. n = 1500 rev/min = 25 rev/ sec$\omega=2\pi\text{n}=2\pi\times25=157.14$
Force of the particle on the blade $\text{Mr}\omega^2=(0.001)\times0.6\times(157.14)=14.8\text{N}$ The fan runs at a full speed in circular path. This exerts the force on the particle (inertia). The particle also exerts a force of 14.8N on the blade along its surface.

View full question & answer→

Question 43 Marks

Find the acceleration of a particle placed on the surface of the earth at the equator due to earth's rotation. The diameter of earth = 12800km and it takes 24 hours for the earth to complete one revolution about its axis.

Answer

Diameter of earth = 12800km Radius $\text{R}=6400\text{km}=64\times10^5\text{m}$$\text{v}=\frac{2\pi\text{R}}{\text{T}}=\frac{2\times3.14\times64\times10^5}{24\times3600}$
$=465.185\text{m}/\text{sec}$
$\text{a}=\frac{\text{V}^2}{\text{R}}=\frac{(46.5185)^2}{64\times10^5}$
$=0.0338\text{m}/\text{sec}^2$

View full question & answer→

Question 53 Marks

In the Bohr model of hydrogen atom, the electron is treated as a particle going in a circle with the centre at the proton. The proton it self is assumed to be fixed in an inertial frame. The centripetal force is provided by the Coloumb attraction. In the ground state, the electron goes round the proton in a circle of radius $5.3 \times 10^{-11}m.$ Find the speed of the electron in the ground state. Mass of the electron $= 9.1 \times 10^{-31}kg$ and charge of the electron $= 1.6 \times 10^{-19}C.$

Answer

Electron revolves around the proton in a circle having proton at the centre.
Centripetal force is provided by coulomb attraction. $r = 5.3 \times 10^{–11}m,$
$m =$ mass of electron $m = 9.1 \times 10^{–31}kg.$
charge of electron $= 1.6 \times 10^{–19}c.$
$\frac{\text{mv}^2}{\text{r}}=\text{k}\frac{\text{q}^2}{\text{r}^2}$
$\Rightarrow\text{v}^2=\frac{\text{kq}^2}{\text{rm}}$
$\Rightarrow\frac{9\times10^9\times1.6\times1.6\times10^{-38}}{5.3\times10^{-11}\times9.1\times10^{-31}}$
$\Rightarrow\frac{23.04}{48.23}\times10^{13}$
$\Rightarrow\text{v}^2=0.477\times10^{13}=4.7\times10^{12}$
$\Rightarrow\text{v}=\sqrt{4.7\times10^{12}}$
$\Rightarrow\text{v}=2.2\times10^6\text{m}/\text{sec}$

View full question & answer→

Question 63 Marks

In a children's park a heavy rod is pivoted at the centre and is made to rotate about the pivot so that the rod always remains horizontal. Two kids hold the rod near the ends and thus rotate with the rod (fig). Let the mass of each kid be 15 kg, the distance between the points of the rod where the two kids hold it be 3.0 m and suppose that the rod rotates at the rate of 20 revolutions per minute. Find the force of friction exerted by the rod on one of the kids.

Answer

d = 3m ⇒ R = 1.5m
R = distance from the centre to one of the kids
N = 20 rev per min $=\frac{20}{60}=\frac{1}{3}\text{rev per sec}$
$\omega=2\pi\text{r}=\frac{2\pi}{3}$
$\text{m}=15\text{kg}$
$\therefore$ Frictional force $\text{F}=\text{mr}\omega^2=15\times(1.5)\times\frac{(2\pi)^2}{9}$
$=5\times(0.5)\times4\pi^2=10\pi^2$
$\therefore$ Frictional force on one of the kids is $10\pi^2$

View full question & answer→

Question 73 Marks

A table with smooth horizontal surface is placed in a cabin which moves in a circle of a large radius R (figure In). A smooth pulley of small radius is fastened to the table. Two masses m and 2m placed on the table are connected through a string going over the pulley. Initially the masses are held by a person with the strings along the outward radius and then the system is released from rest (with respect to the cabin). Find the magnitude of the initial acceleration of the masses as seen from the cabin and the tension in the string.

Answer

Let the bigger mass accelerates towards right with ‘a’.
From the free body diagrams,
$\text{T}-\text{ma}-\text{m}\omega^2\text{R}=0\ \dots(1)$
$\text{T}+2\text{ma}-2\text{m}\omega^2\text{R}=0\ \dots(2)$
Eq (1) – Eq (2)
$\Rightarrow3\text{ma}=\text{m}\omega^2\text{R}$
$\Rightarrow\text{a}=\frac{\text{m}\omega^2\text{R}}{3}$
Substituting the value of a in Equation (1), we get $\text{T}=\frac{4}{3}\text{m}\omega^2\text{R}.$

View full question & answer→

Question 83 Marks

Consider the circular motion of the earth around the sun. Which of the following statements is more appropriate?

Gravitational attraction of the sun on the earth is equal to the centripetal force.
Gravitational attraction of the sun on the earth is the centripetal force.

Answer

Gravitational attraction is equal to centripetal force. as both forces are different forces so they cannot be same. but these forces can be equal.
In case Gravitational attraction would more than centripetal force earth would move closer to sun. in case Gravitational attraction is lesser than Centripetal force then earth would move away from sun.

View full question & answer→

Question 93 Marks

Suppose the bob of the previous problem has a speed of 1.4m/ s when the string makes an angle of 0.20 radian with the vertical. Find the tension at this instant. You can use $\cos\theta\approx1-\frac{\theta^2}{2}$ and $\sin\theta\approx\theta$ for small $\theta.$

Answer

Bob has a velocity 1.4m/ sec, when the string makes an angle of 0.2 radian.
m = 100g = 0.1kg, r = 1m, v = 1.4m/ sec.
From the diagram,
$\text{T}-\text{mg}\cos\theta=\frac{\text{mv}^2}{\text{R}}$
$\Rightarrow\text{T}=\frac{\text{mv}^2}{\text{R}}+\text{mg}\cos\theta$
$\Rightarrow\text{T}=\frac{0.1\times(1.4)^2}{1}+(0.1)\times9.8\times\Big(1-\frac{\theta^2}{2}\Big)$
$\Rightarrow\text{T}=0.196+9.8\times\Big(1-\frac{(0.2)^2}{2}\Big)$ $\Big(\therefore\ \cos\theta=1-\frac{\theta^2}{2}\text{for small}\ \theta\Big)$
$\Rightarrow\text{T}=0.196+(0.98)\times(0.98)$
$\Rightarrow\text{T}=0.196+0.964$
$\Rightarrow\text{T}=1.156\text{N}\approx1.16\text{N}$

View full question & answer→

Question 103 Marks

A particle moves in a circle of radius 1.0cm at a speed given by v = 2.0t where v is in cm/s and t in seconds.

Find the radial acceleration of the particle at t = 1s.
Find the tangential acceleration at t = 1s.
Find the magnitude of the acceleration at t = 1s.

Answer

V = 2t, r = 1cm

Radial acceleration at t = 1sec.

$\text{a}=\frac{\text{v}^2}{\text{r}}=\frac{2^2}{1}$
$=4\text{cm}/\text{sec}^2$

Tangential acceleration at t = 1sec.

$\text{a}=\frac{\text{dv}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(2\text{t})$
$=2\text{cm}/\text{sec}^2$

Magnitude of acceleration at t = 1sec.

$\text{a}=\sqrt{4^2+2^2}$
$=\sqrt{20}\text{cm}/\text{sec}^2$

View full question & answer→

Question 113 Marks

A mosquito is sitting on an $L.P.$ record disc rotating on a turn table at $33\frac{1}{3}$ revolutions per minute. The distance of the mosquito from the centre of the turn table is $10\ cm$. Show that the friction coefficient between the record and the mosquito is greater than $\frac{\pi^2}{81}.$ Take $g =10 m/s^2.$

Answer

A mosquito is sitting on an $L.P.$ record disc $\&$ rotating on a turn table at $33\frac{1}{3}$ rpm.$\text{n}=33\frac{1}{3}\text{rpm}=\frac{100}{3\times60}\text{rps}$
$\therefore\ \omega=2\pi\text{n}=2\pi\times\frac{100}{180}$
$\frac{10\pi}{9}\text{rad}/\text{sec}$
$\text{r}=10\text{cm}=0.1\text{m},\ \text{g}=10\text{m}/\text{sec}^2$
$\mu\text{mg}\geq\text{mr}\omega^2$
$\Rightarrow\mu=\frac{\text{r}\omega^2}{\text{g}}\geq\frac{0.1\times\Big(\frac{10\pi}{9}\Big)^2}{10}$
$\Rightarrow\mu\geq\frac{\pi^2}{81}$

View full question & answer→

Question 123 Marks

What is the radius of curvature of the parabola traced out by the projectile in the previous problem at a point where the particle velocity makes an angle $\frac{\theta}{2}$ with the horizontal?

Answer

Let ‘u’ the velocity at the pt where it makes an angle $\frac{\theta}{2}$ with horizontal. The horizontal component remains unchanged So, $\text{v}\cos\frac{\theta}{2}=\omega\cos\theta$$\Rightarrow\text{v}=\frac{\text{u}\cos\theta}{\cos\Big(\frac{\theta}{2}\Big)}\ \dots(1)$
From figure

$\text{mg}\cos\Big(\frac{\theta}{2}\Big)=\frac{\text{mv}^2}{\text{r}}$
$\Rightarrow\text{r}=\frac{\text{v}^2}{\text{g}\cos\Big(\frac{\theta}{2}\Big)}$
putting the value of ‘v’ from equn(1)$\text{r}=\frac{\text{u}^2\cos^2\theta}{\text{g}\cos^3\Big(\frac{\theta}{2}\Big)}$

View full question & answer→

Question 133 Marks

A particle is projected with a speed u at an angle $\theta$ with the horizontal. Consider a small part of its path near the highest position and take it approximately to be a circular arc. What is the radius of this circle? This radius is called the radius of curvature of the curve at the point.

Answer

Particle is projected with speed ‘u’ at an angle $\theta.$ At the highest pt. the vertical component of velocity is‘0’.
So, at that point, velocity $=\text{u}\sin\theta$
centripetal force $=\text{mu}^2\cos^2\Big(\frac{\theta}{\text{r}}\Big)$
At highest pt.
$\text{mg}=\frac{\text{mv}^2}{\text{r}}$
$\Rightarrow\text{r}=\frac{\text{u}^2\cos^2\theta}{\text{g}}$

View full question & answer→

Physics 3 Marks Question

Test yourself on this topic