Q. No.The height in terms of radius of the earth \((R)\), at which the acceleration due to gravity becomes \(\dfrac{g}{9},\) where \(g\) is acceleration due to gravity on earth's surface, is:
1. \(\sqrt{3}{R}\)
2. \({2}\sqrt{2}{R}\)
3. \(2R\)
4. \(\dfrac{4}{9}R\)
1. \(\sqrt{3}{R}\)
2. \({2}\sqrt{2}{R}\)
3. \(2R\)
4. \(\dfrac{4}{9}R\)
Subtopic: Acceleration due to Gravity |
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When one moves from a point \(16~\text{km}\) below the earth's surface to a point \(16~\text{km}\) above the earth's surface. The change in \(g\) is approximately \(\alpha \% \). The value of \(\alpha\) is:
1. \(0.12\)
2. \(0.25\)
3. \(0.50\)
4. \(0.75\)
1. \(0.12\)
2. \(0.25\)
3. \(0.50\)
4. \(0.75\)
Subtopic: Acceleration due to Gravity |
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Acceleration due to gravity on the surface of earth is \(g.\) If the diameter of earth is reduced to one third of its original value and mass remains unchanged, then the acceleration due to gravity on the surface of the earth is _______________ \(g.\)
1. \(9\)
2. \(18\)
3. \(27\)
4. \(20\)
1. \(9\)
2. \(18\)
3. \(27\)
4. \(20\)
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A \(90\) kg body placed at \(2R\) distance from surface of earth experiences gravitational pull of (\(R=\) Radius of earth, \(g=10\) ms-2)
1. \(100\) N
2. \(300\) N
3. \(120\) N
4. \(225\) N
1. \(100\) N
2. \(300\) N
3. \(120\) N
4. \(225\) N
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Assuming the earth to be a sphere of uniform mass density, a body weighed \(300\) N on the surface of earth. How much it would weight at \(R/4\) depth under the surface of the earth?
1. \(300\) N
2. \(75\) N
3. \(225\) N
4. \(375\) NSubtopic: Acceleration due to Gravity |
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If an object has the same weight at the same distance above and below the surface of the earth, then its distance from the surface of the earth is:
1. \(\frac{R}{2}\)
2. \((\sqrt{5}-1) \frac{R}{2}\)
3. \((\sqrt{3}-1) \frac{R}{2}\)
4. \((\sqrt{5}-1) R\)
1. \(\frac{R}{2}\)
2. \((\sqrt{5}-1) \frac{R}{2}\)
3. \((\sqrt{3}-1) \frac{R}{2}\)
4. \((\sqrt{5}-1) R\)
Subtopic: Acceleration due to Gravity |
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If the diameter of the earth becomes half keeping mass constant, then the acceleration due to gravity at the surface of the earth becomes:
| 1. | half | 2. | four times |
| 3. | twice | 4. | three times |
Subtopic: Acceleration due to Gravity |
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On a planet \(\rho\) (mass density) is same as that of earth while mass of planet is twice than that of earth. Ratio of weight of a body on surface of planet to that on earth is equal to:
1. 1
2. (2)1/3
3. (2)-1/3
4. 2
1. 1
2. (2)1/3
3. (2)-1/3
4. 2
Subtopic: Acceleration due to Gravity |
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An object weighs 200 \(N\) at the surface of earth. Find the weight at a depth of \(R\over 2\), where \(R\) is the radius of earth.
1. 100 \(N\)
2. 300 \(N\)
3. 50 \(N\)
4. 150 \(N\)
1. 100 \(N\)
2. 300 \(N\)
3. 50 \(N\)
4. 150 \(N\)
Subtopic: Acceleration due to Gravity |
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Which of the following expressions gives the value of acceleration due to gravity \((g')\) at the altitude \(h\) above the surface of the Earth?
(\(R:\) radius of Earth; \(g:\) acceleration due to gravity at the surface of Earth)
1. \(g^{\prime}=g \dfrac{h^2}{R^2} \)
2. \(g^{\prime}=\dfrac{g R^2}{(R+h)^2} \)
3. \(g^{\prime}=g\left(1-\dfrac{h}{R}\right) \)
4. \(g^{\prime}=g\left(1-\dfrac{h^2}{R^2}\right)\)
(\(R:\) radius of Earth; \(g:\) acceleration due to gravity at the surface of Earth)
1. \(g^{\prime}=g \dfrac{h^2}{R^2} \)
2. \(g^{\prime}=\dfrac{g R^2}{(R+h)^2} \)
3. \(g^{\prime}=g\left(1-\dfrac{h}{R}\right) \)
4. \(g^{\prime}=g\left(1-\dfrac{h^2}{R^2}\right)\)
Subtopic: Acceleration due to Gravity |
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