NEET-XII-Physics
11: Gravitation
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- #20-aShow that the equation is dimensionally correct.Ans : `` V=\left(\frac{20\,\mathrm{\,N\,}}{\,\mathrm{\,kg\,}}\right)\left(x+y\right)``
`` \left[\frac{\,\mathrm{\,GM\,}}{\,\mathrm{\,R\,}}\right]=\left[\frac{{\,\mathrm{\,MLT\,}}^{-2}}{\,\mathrm{\,M\,}}\right]\left[\,\mathrm{\,L\,}\right]``
`` \Rightarrow \left[\frac{{\,\mathrm{\,M\,}}^{-1}{\,\mathrm{\,L\,}}^{3}{\,\mathrm{\,T\,}}^{-2}{\,\mathrm{\,M\,}}^{1}}{\,\mathrm{\,L\,}}\right]=\left[\frac{{\,\mathrm{\,ML\,}}^{2}{\,\mathrm{\,T\,}}^{-2}}{\,\mathrm{\,M\,}}\right]``
`` \Rightarrow \left[{\,\mathrm{\,M\,}}^{0}{\,\mathrm{\,L\,}}^{2}{\,\mathrm{\,T\,}}^{-2}\right]=\left[{\,\mathrm{\,M\,}}^{0}{\,\mathrm{\,L\,}}^{2}{\,\mathrm{\,T\,}}^{-2}\right]``
`` \therefore \,\mathrm{\,LHS\,}=\,\mathrm{\,RHS\,}``
- #20-bFind the gravitational field at the point (x, y). Leave your answer in terms of the unit vectors
i→, j→, k→.Ans : The gravitational field at the point (x, y) is given by `` {\stackrel{\to }{\,\mathrm{\,E\,}}}_{\left(x,y\right)}=-20\left(\frac{\,\mathrm{\,N\,}}{\,\mathrm{\,kg\,}}\right)\stackrel{‸}{i}-\left(\frac{20\,\mathrm{\,N\,}}{\,\mathrm{\,kg\,}}\right)\stackrel{‸}{j}``.
- #20-cCalculate the magnitude of the gravitational force on a particle of mass 500 g placed at the origin.Ans : `` \stackrel{\to }{\,\mathrm{\,F\,}}=\stackrel{\to }{\,\mathrm{\,E\,}}m``
`` =0.5\,\mathrm{\,kg\,}\left[-\left(\frac{20\,\mathrm{\,N\,}}{\,\mathrm{\,kg\,}}\right)\stackrel{‸}{i}-\left(\frac{20\,\mathrm{\,N\,}}{\,\mathrm{\,kg\,}}\right)\stackrel{‸}{j}\right]``
`` =-\left(10\,\mathrm{\,N\,}\right)\stackrel{‸}{i}-\left(10\,\mathrm{\,N\,}\right)\stackrel{‸}{j}``
`` \therefore \left|\stackrel{\to }{\,\mathrm{\,F\,}}\right|=\sqrt{\left(100\right)+\left(100\right)}``
`` =10\sqrt{2}\,\mathrm{\,N\,}``
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- Qstn #21The gravitational field in a region is given by
E=2i→+3j→ N kg-1. Show that no work is done by the gravitational field when a particle is moved on the line 3y + 2x = 5.
[
Hint : If a line y = mx + c makes angle θ with the X-axis, m = tan θ.]Ans : The gravitational field in a region is given by `` \stackrel{\to }{\,\mathrm{\,E\,}}=2\stackrel{‸}{i}+3\stackrel{‸}{j}``.
Slope of the electric field, `` {m}_{1}=\,\mathrm{\,tan\,}{\,\mathrm{\,\theta \,}}_{1}=\frac{3}{2}``
The given line is 3y + 2x = 5.
Slope of the line, `` {m}_{2}=\,\mathrm{\,tan\,}{\,\mathrm{\,\theta \,}}_{2}=-\frac{2}{3}``
We can see that m1m2 = -1
Since the directions of the field and the displacement are perpendicular to earth other, no work is done by the gravitational field when a particle is moved on the given line.
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- Qstn #22Find the height over the Earth’s surface at which the weight of a body becomes half of its value at the surface.Ans : Assume that at height h, the weight of the body becomes half.
Weight of the body at the surface = mg
Weight of the body at height h above the Earth's surface = mg', where g' is the acceleration due to gravity at height h
`` \,\mathrm{\,Now\,},g\text{'}=\frac{1}{2}g``
`` \therefore \left(\frac{1}{2}\right)\frac{\,\mathrm{\,G\,}M}{{R}^{2}}=\frac{\,\mathrm{\,G\,}M}{{\left(R+h\right)}^{2}}\left[\because g=\frac{\,\mathrm{\,G\,}\,\mathrm{\,M\,}}{{\,\mathrm{\,R\,}}^{2}}\right]``
`` \Rightarrow 2{R}^{2}={\left(R+h\right)}^{2}``
`` \Rightarrow \sqrt{2}R=R+h``
`` \Rightarrow h=\left(\sqrt{2}-1\right)R``
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- Qstn #23What is the acceleration due to gravity on the top of Mount Everest? Mount Everest is the highest mountain peak of the world at the height of 8848 m. The value at sea level is 9.80 m s-2.Ans : Let g' be the acceleration due to gravity on Mount Everest.
`` \,\mathrm{\,Then\,}g\text{'}=g\left(1-\frac{2h}{R}\right)``
`` \,\mathrm{\,here\,}h=8848\,\mathrm{\,m\,}``
`` =9.8\left(1-0.00276\right)``
`` =9.0\left(0.99724\right)``
`` =9.77\,\mathrm{\,m\,}/{\,\mathrm{\,s\,}}^{2}``
∴ The acceleration due to gravity on the top of Mount Everest is 9.77 m/s2.
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- Qstn #24Find the acceleration due to gravity in a mine of depth 640 m if the value at the surface is 9.800 m s-2. The radius of the earth is 6400 km.Ans : Let g' be the acceleration due to gravity in a mine of depth d.
`` \therefore g\text{'}=g\left(1-\frac{d}{R}\right)``
`` =9.8\left(1-\frac{640}{640\times {10}^{3}}\right)``
`` =9.8\left(\frac{10000-1}{{10}^{4}}\right)``
`` =\frac{9.8}{{10}^{4}}\times 9999``
`` =9.8\times 0.9999``
`` =9.799\,\mathrm{\,m\,}/{\,\mathrm{\,s\,}}^{2}``
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- Qstn #25A body is weighed by a spring balance to be 1.000 kg at the North Pole. How much will it weigh at the equator? Account for the earth’s rotation only.Ans : Let gp be the acceleration due to gravity at the poles.
Let ge be the acceleration due to gravity at the equator.
Now, acceleration due to gravity at the equator is given by
ge = gp `` -`` ω2r
= 9.81 - (7.3 × 10-5)2 × 6400 × 103
= 9.81 - (53.29 × 10-10) × 64 × 105
= 9.81 - 0.034 = 9.776 m/s2
Now, mge = 1 kg × 9.776 m/s2
= 9.776 N
∴ The body will weigh 9.776 N at the equator.
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- Qstn #26A body stretches a spring by a particular length at the earth’s surface at the equator. At what height above the south pole will it stretch the same spring by the same length? Assume the earth to be spherical.Ans : At the equator, g' = g - ω2R ...(i)
Let h be the height above the South Pole where the body stretch the spring by the same length.
The acceleration due to gravity at this point is `` g\text{'}=g\left(1-\frac{2h}{\,\mathrm{\,R\,}}\right)``.
Weight of the body at the equator = weight of the body at height h above the South Pole
`` \therefore g-{\,\mathrm{\,\omega \,}}^{2}r=g\left(1-\left(\frac{2h}{R}\right)\right)``
`` \Rightarrow 1-\frac{{\,\mathrm{\,\omega \,}}^{2}\,\mathrm{\,R\,}}{2g}=1-\frac{2h}{R}``
`` \Rightarrow h=\frac{{\,\mathrm{\,\omega \,}}^{2}{R}^{2}}{2g}``
`` =\frac{{\left(7.3\times {10}^{-5}\right)}^{2}\times {\left(6400\times {10}^{-3}\right)}^{2}}{2\times 9.81}``
`` =\frac{{\left(7.3\right)}^{2}\times {\left(64\right)}^{2}}{19.62}``
`` =11125\,\mathrm{\,m\,}=10\,\mathrm{\,km\,}\left(\,\mathrm{\,approx\,}.\right)``
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- Qstn #27At what rate should the earth rotate so that the apparent g at the equator becomes zero? What will be the length of the day in this situation?Ans : The apparent acceleration due to gravity at the equator becomes zero.
i.e., g' = g - ω2R = 0
⇒ g = ω2R
`` \Rightarrow \,\mathrm{\,\omega \,}=\sqrt{\frac{g}{R}}=\sqrt{\frac{9.8}{6400\times {10}^{3}}}``
`` \Rightarrow \,\mathrm{\,\omega \,}=\sqrt{\frac{9.8\times {10}^{-5}}{6.4}}=\sqrt{1.5\times {10}^{-6}}``
`` \Rightarrow \,\mathrm{\,\omega \,}=1.2\times {10}^{-3}\,\mathrm{\,rad\,}/\,\mathrm{\,s\,}``
`` \therefore T=\frac{2\,\mathrm{\,\pi \,}}{\,\mathrm{\,\omega \,}}=\frac{2\times 3.14}{1.2\times {10}^{-3}}``
`` =\frac{6.28}{1.2\times {10}^{-3}}``
`` =1.41\,\mathrm{\,h\,}``
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- Qstn #28A pendulum having a bob of mass m is hanging in a ship sailing along the equator from east to west. When the ship is stationary with respect to water the tension in the string is T0.
- #28-aFind the speed of the ship due to rotation of the earth about its axis.Ans : Speed of the ship due to rotation of the Earth is v = ωR, where R is the radius of the Earth and ω is its angular speed.
- #28-bFind the difference between T0 and the earth’s attraction on the bob.Ans : The tension in the string is given by
T0 = mg - mω2R
∴ T0 - mg = mω2R
- #28-cIf the ship sails at speed v, what is the tension in the string? Angular speed of earth’s rotation is ω and radius of the earth is R.Ans : Let the ship move with a speed v.
Then the tension in the string is given by
`` T=mg-m{\,\mathrm{\,\omega \,}}_{1}^{2}R``
`` ={T}_{0}-\frac{{\left(v-\omega R\right)}^{2}}{{R}^{2}}R``
`` ={T}_{0}-\frac{\left({v}^{2}+{\,\mathrm{\,\omega \,}}^{2}{R}^{2}-2\omega Rv\right)}{R}R``
`` \therefore T={T}_{0}+2\,\mathrm{\,\omega \,}vm``
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