163994
The value of ' \(g\) ' at a place is \(9.8 \mathrm{~ms}^{-2}\), if the earth shrinks uniformly (without changing its mass or centre) to half its size the value of ' \(g\) ' at the same point will now be :
1 \(9.8 \mathrm{~m} \mathrm{~s}^{-2}\)
2 \(19.6 \mathrm{~m} \mathrm{~s}^{-2}\)
3 \(4.9 \mathrm{~m} \mathrm{~s}^{-2}\)
4 zero
Explanation:
Since the point remains the same \(\therefore \mathrm{M}\) and \(\mathrm{R}\) are same and \(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2}\) remains the same
NCERT-XI- I -133
5 RBTS PAPER
163995
The weight of the body at the centre of earth is :
1 zero
2 infinity
3 9.8 times the weight on the surface
4 None of these
Explanation:
The weight of the body is \(W=m g\). At centre of earth the value of \(g\) becomes zero So weight becomes zero at centre of earth
NCERT-XI- I -134
5 RBTS PAPER
163996
To what height the body is to be raised so that force of gravity on it is \(1 \%\) of the value on the surfaace of earth? (Radius of earth is \(6400 \mathrm{~km}\) ) :
1 \(6400 \mathrm{~km}\)
2 \(12800 \mathrm{~km}\)
3 \(57600 \mathrm{~km}\)
4 \(3200 \mathrm{~km}\)
Explanation:
\(\frac{\mathrm{mg}^{\prime}}{\mathrm{mg}}=\frac{1}{100}\) or \(\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{1}{100}=\frac{(\mathrm{R})^2}{(\mathrm{R}+\mathrm{h})^2}\) or \(\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}=\frac{1}{10}\) or \(9 \mathrm{R}=\mathrm{h}\) or \(h=6400 \times 9=57600 \mathrm{~km}\)
NCERT-XI- I -133
5 RBTS PAPER
163997
The period of satellite in a circular orbit of radius \(\mathbf{R}_0\) is \(\mathrm{T}\), the period of another satellite in circular orbit of radius \(4 \mathbf{R}_0\) is :
1 \(T / 4\)
2 \(T / 8\)
3 \(4 \mathrm{~T}\)
4 \(8 \mathrm{~T}\)
Explanation:
\(\mathrm{T} \propto \mathrm{R}_0^{3 / 2}\) and \(\left(4 \mathrm{R}_0\right)^{3 / 2}\) or \(\frac{T^{\prime}}{T}=(4)^{3 / 2}=(64)^{1 / 2}=8\)
NCERT-XI- I -137
5 RBTS PAPER
163998
The time period of simple pendulum in a satellite is :
163994
The value of ' \(g\) ' at a place is \(9.8 \mathrm{~ms}^{-2}\), if the earth shrinks uniformly (without changing its mass or centre) to half its size the value of ' \(g\) ' at the same point will now be :
1 \(9.8 \mathrm{~m} \mathrm{~s}^{-2}\)
2 \(19.6 \mathrm{~m} \mathrm{~s}^{-2}\)
3 \(4.9 \mathrm{~m} \mathrm{~s}^{-2}\)
4 zero
Explanation:
Since the point remains the same \(\therefore \mathrm{M}\) and \(\mathrm{R}\) are same and \(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2}\) remains the same
NCERT-XI- I -133
5 RBTS PAPER
163995
The weight of the body at the centre of earth is :
1 zero
2 infinity
3 9.8 times the weight on the surface
4 None of these
Explanation:
The weight of the body is \(W=m g\). At centre of earth the value of \(g\) becomes zero So weight becomes zero at centre of earth
NCERT-XI- I -134
5 RBTS PAPER
163996
To what height the body is to be raised so that force of gravity on it is \(1 \%\) of the value on the surfaace of earth? (Radius of earth is \(6400 \mathrm{~km}\) ) :
1 \(6400 \mathrm{~km}\)
2 \(12800 \mathrm{~km}\)
3 \(57600 \mathrm{~km}\)
4 \(3200 \mathrm{~km}\)
Explanation:
\(\frac{\mathrm{mg}^{\prime}}{\mathrm{mg}}=\frac{1}{100}\) or \(\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{1}{100}=\frac{(\mathrm{R})^2}{(\mathrm{R}+\mathrm{h})^2}\) or \(\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}=\frac{1}{10}\) or \(9 \mathrm{R}=\mathrm{h}\) or \(h=6400 \times 9=57600 \mathrm{~km}\)
NCERT-XI- I -133
5 RBTS PAPER
163997
The period of satellite in a circular orbit of radius \(\mathbf{R}_0\) is \(\mathrm{T}\), the period of another satellite in circular orbit of radius \(4 \mathbf{R}_0\) is :
1 \(T / 4\)
2 \(T / 8\)
3 \(4 \mathrm{~T}\)
4 \(8 \mathrm{~T}\)
Explanation:
\(\mathrm{T} \propto \mathrm{R}_0^{3 / 2}\) and \(\left(4 \mathrm{R}_0\right)^{3 / 2}\) or \(\frac{T^{\prime}}{T}=(4)^{3 / 2}=(64)^{1 / 2}=8\)
NCERT-XI- I -137
5 RBTS PAPER
163998
The time period of simple pendulum in a satellite is :
163994
The value of ' \(g\) ' at a place is \(9.8 \mathrm{~ms}^{-2}\), if the earth shrinks uniformly (without changing its mass or centre) to half its size the value of ' \(g\) ' at the same point will now be :
1 \(9.8 \mathrm{~m} \mathrm{~s}^{-2}\)
2 \(19.6 \mathrm{~m} \mathrm{~s}^{-2}\)
3 \(4.9 \mathrm{~m} \mathrm{~s}^{-2}\)
4 zero
Explanation:
Since the point remains the same \(\therefore \mathrm{M}\) and \(\mathrm{R}\) are same and \(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2}\) remains the same
NCERT-XI- I -133
5 RBTS PAPER
163995
The weight of the body at the centre of earth is :
1 zero
2 infinity
3 9.8 times the weight on the surface
4 None of these
Explanation:
The weight of the body is \(W=m g\). At centre of earth the value of \(g\) becomes zero So weight becomes zero at centre of earth
NCERT-XI- I -134
5 RBTS PAPER
163996
To what height the body is to be raised so that force of gravity on it is \(1 \%\) of the value on the surfaace of earth? (Radius of earth is \(6400 \mathrm{~km}\) ) :
1 \(6400 \mathrm{~km}\)
2 \(12800 \mathrm{~km}\)
3 \(57600 \mathrm{~km}\)
4 \(3200 \mathrm{~km}\)
Explanation:
\(\frac{\mathrm{mg}^{\prime}}{\mathrm{mg}}=\frac{1}{100}\) or \(\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{1}{100}=\frac{(\mathrm{R})^2}{(\mathrm{R}+\mathrm{h})^2}\) or \(\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}=\frac{1}{10}\) or \(9 \mathrm{R}=\mathrm{h}\) or \(h=6400 \times 9=57600 \mathrm{~km}\)
NCERT-XI- I -133
5 RBTS PAPER
163997
The period of satellite in a circular orbit of radius \(\mathbf{R}_0\) is \(\mathrm{T}\), the period of another satellite in circular orbit of radius \(4 \mathbf{R}_0\) is :
1 \(T / 4\)
2 \(T / 8\)
3 \(4 \mathrm{~T}\)
4 \(8 \mathrm{~T}\)
Explanation:
\(\mathrm{T} \propto \mathrm{R}_0^{3 / 2}\) and \(\left(4 \mathrm{R}_0\right)^{3 / 2}\) or \(\frac{T^{\prime}}{T}=(4)^{3 / 2}=(64)^{1 / 2}=8\)
NCERT-XI- I -137
5 RBTS PAPER
163998
The time period of simple pendulum in a satellite is :
163994
The value of ' \(g\) ' at a place is \(9.8 \mathrm{~ms}^{-2}\), if the earth shrinks uniformly (without changing its mass or centre) to half its size the value of ' \(g\) ' at the same point will now be :
1 \(9.8 \mathrm{~m} \mathrm{~s}^{-2}\)
2 \(19.6 \mathrm{~m} \mathrm{~s}^{-2}\)
3 \(4.9 \mathrm{~m} \mathrm{~s}^{-2}\)
4 zero
Explanation:
Since the point remains the same \(\therefore \mathrm{M}\) and \(\mathrm{R}\) are same and \(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2}\) remains the same
NCERT-XI- I -133
5 RBTS PAPER
163995
The weight of the body at the centre of earth is :
1 zero
2 infinity
3 9.8 times the weight on the surface
4 None of these
Explanation:
The weight of the body is \(W=m g\). At centre of earth the value of \(g\) becomes zero So weight becomes zero at centre of earth
NCERT-XI- I -134
5 RBTS PAPER
163996
To what height the body is to be raised so that force of gravity on it is \(1 \%\) of the value on the surfaace of earth? (Radius of earth is \(6400 \mathrm{~km}\) ) :
1 \(6400 \mathrm{~km}\)
2 \(12800 \mathrm{~km}\)
3 \(57600 \mathrm{~km}\)
4 \(3200 \mathrm{~km}\)
Explanation:
\(\frac{\mathrm{mg}^{\prime}}{\mathrm{mg}}=\frac{1}{100}\) or \(\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{1}{100}=\frac{(\mathrm{R})^2}{(\mathrm{R}+\mathrm{h})^2}\) or \(\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}=\frac{1}{10}\) or \(9 \mathrm{R}=\mathrm{h}\) or \(h=6400 \times 9=57600 \mathrm{~km}\)
NCERT-XI- I -133
5 RBTS PAPER
163997
The period of satellite in a circular orbit of radius \(\mathbf{R}_0\) is \(\mathrm{T}\), the period of another satellite in circular orbit of radius \(4 \mathbf{R}_0\) is :
1 \(T / 4\)
2 \(T / 8\)
3 \(4 \mathrm{~T}\)
4 \(8 \mathrm{~T}\)
Explanation:
\(\mathrm{T} \propto \mathrm{R}_0^{3 / 2}\) and \(\left(4 \mathrm{R}_0\right)^{3 / 2}\) or \(\frac{T^{\prime}}{T}=(4)^{3 / 2}=(64)^{1 / 2}=8\)
NCERT-XI- I -137
5 RBTS PAPER
163998
The time period of simple pendulum in a satellite is :
163994
The value of ' \(g\) ' at a place is \(9.8 \mathrm{~ms}^{-2}\), if the earth shrinks uniformly (without changing its mass or centre) to half its size the value of ' \(g\) ' at the same point will now be :
1 \(9.8 \mathrm{~m} \mathrm{~s}^{-2}\)
2 \(19.6 \mathrm{~m} \mathrm{~s}^{-2}\)
3 \(4.9 \mathrm{~m} \mathrm{~s}^{-2}\)
4 zero
Explanation:
Since the point remains the same \(\therefore \mathrm{M}\) and \(\mathrm{R}\) are same and \(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2}\) remains the same
NCERT-XI- I -133
5 RBTS PAPER
163995
The weight of the body at the centre of earth is :
1 zero
2 infinity
3 9.8 times the weight on the surface
4 None of these
Explanation:
The weight of the body is \(W=m g\). At centre of earth the value of \(g\) becomes zero So weight becomes zero at centre of earth
NCERT-XI- I -134
5 RBTS PAPER
163996
To what height the body is to be raised so that force of gravity on it is \(1 \%\) of the value on the surfaace of earth? (Radius of earth is \(6400 \mathrm{~km}\) ) :
1 \(6400 \mathrm{~km}\)
2 \(12800 \mathrm{~km}\)
3 \(57600 \mathrm{~km}\)
4 \(3200 \mathrm{~km}\)
Explanation:
\(\frac{\mathrm{mg}^{\prime}}{\mathrm{mg}}=\frac{1}{100}\) or \(\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{1}{100}=\frac{(\mathrm{R})^2}{(\mathrm{R}+\mathrm{h})^2}\) or \(\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}=\frac{1}{10}\) or \(9 \mathrm{R}=\mathrm{h}\) or \(h=6400 \times 9=57600 \mathrm{~km}\)
NCERT-XI- I -133
5 RBTS PAPER
163997
The period of satellite in a circular orbit of radius \(\mathbf{R}_0\) is \(\mathrm{T}\), the period of another satellite in circular orbit of radius \(4 \mathbf{R}_0\) is :
1 \(T / 4\)
2 \(T / 8\)
3 \(4 \mathrm{~T}\)
4 \(8 \mathrm{~T}\)
Explanation:
\(\mathrm{T} \propto \mathrm{R}_0^{3 / 2}\) and \(\left(4 \mathrm{R}_0\right)^{3 / 2}\) or \(\frac{T^{\prime}}{T}=(4)^{3 / 2}=(64)^{1 / 2}=8\)
NCERT-XI- I -137
5 RBTS PAPER
163998
The time period of simple pendulum in a satellite is :
