NEET Test Series from KOTA - 10 Papers In MS WORD
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Gravitation
270600
The height at which the value of acceleration due to gravity becomes \(50 \%\) of that at the surface of the earth. \((\) Radius of the earth \(=\) \(6400 \mathrm{~km})\) is
270601
The radius and density of two artificial satellites are \(R_{1}, R_{2}\) and \(\rho_{1}, \rho_{2}\) respectively. The ratio of acceleration due to gravities on them will be
1 \(\frac{R_{2} \rho_{2}}{R_{1} \rho_{1}}\)
2 \(\frac{R_{1} \rho_{2}}{R_{2} \rho_{1}}\)
3 \(\frac{R_{1} \rho_{1}}{R_{2} \rho_{2}}\)
4 \(\frac{R_{2} \rho_{1}}{R_{1} \rho_{2}}\)
Explanation:
\(g=\frac{4}{3} \pi G R \rho \Rightarrow g \propto R \rho\)
Gravitation
270602
A man weigh ' \(W\) ' on the surface of earth and his weight at a height ' \(R\) ' from surface of earth is ( \(R\) is Radius of earth)
1 \(\frac{W}{4}\)
2 \(\frac{W}{2}\)
3 \(\mathrm{W}\)
4 \(4 \mathrm{~W}\)
Explanation:
\(W=m g ; W^{\prime}=m g^{\prime}=m g \because \frac{R}{R+h} \overbrace{}^{2}=\frac{W}{4}\)
Gravitation
270603
The acceleration due to gravity at the latitude \(45^{\circ}\) on the earth becomes zero if the angular velocity of rotation of earth is
270600
The height at which the value of acceleration due to gravity becomes \(50 \%\) of that at the surface of the earth. \((\) Radius of the earth \(=\) \(6400 \mathrm{~km})\) is
270601
The radius and density of two artificial satellites are \(R_{1}, R_{2}\) and \(\rho_{1}, \rho_{2}\) respectively. The ratio of acceleration due to gravities on them will be
1 \(\frac{R_{2} \rho_{2}}{R_{1} \rho_{1}}\)
2 \(\frac{R_{1} \rho_{2}}{R_{2} \rho_{1}}\)
3 \(\frac{R_{1} \rho_{1}}{R_{2} \rho_{2}}\)
4 \(\frac{R_{2} \rho_{1}}{R_{1} \rho_{2}}\)
Explanation:
\(g=\frac{4}{3} \pi G R \rho \Rightarrow g \propto R \rho\)
Gravitation
270602
A man weigh ' \(W\) ' on the surface of earth and his weight at a height ' \(R\) ' from surface of earth is ( \(R\) is Radius of earth)
1 \(\frac{W}{4}\)
2 \(\frac{W}{2}\)
3 \(\mathrm{W}\)
4 \(4 \mathrm{~W}\)
Explanation:
\(W=m g ; W^{\prime}=m g^{\prime}=m g \because \frac{R}{R+h} \overbrace{}^{2}=\frac{W}{4}\)
Gravitation
270603
The acceleration due to gravity at the latitude \(45^{\circ}\) on the earth becomes zero if the angular velocity of rotation of earth is
270600
The height at which the value of acceleration due to gravity becomes \(50 \%\) of that at the surface of the earth. \((\) Radius of the earth \(=\) \(6400 \mathrm{~km})\) is
270601
The radius and density of two artificial satellites are \(R_{1}, R_{2}\) and \(\rho_{1}, \rho_{2}\) respectively. The ratio of acceleration due to gravities on them will be
1 \(\frac{R_{2} \rho_{2}}{R_{1} \rho_{1}}\)
2 \(\frac{R_{1} \rho_{2}}{R_{2} \rho_{1}}\)
3 \(\frac{R_{1} \rho_{1}}{R_{2} \rho_{2}}\)
4 \(\frac{R_{2} \rho_{1}}{R_{1} \rho_{2}}\)
Explanation:
\(g=\frac{4}{3} \pi G R \rho \Rightarrow g \propto R \rho\)
Gravitation
270602
A man weigh ' \(W\) ' on the surface of earth and his weight at a height ' \(R\) ' from surface of earth is ( \(R\) is Radius of earth)
1 \(\frac{W}{4}\)
2 \(\frac{W}{2}\)
3 \(\mathrm{W}\)
4 \(4 \mathrm{~W}\)
Explanation:
\(W=m g ; W^{\prime}=m g^{\prime}=m g \because \frac{R}{R+h} \overbrace{}^{2}=\frac{W}{4}\)
Gravitation
270603
The acceleration due to gravity at the latitude \(45^{\circ}\) on the earth becomes zero if the angular velocity of rotation of earth is
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Gravitation
270600
The height at which the value of acceleration due to gravity becomes \(50 \%\) of that at the surface of the earth. \((\) Radius of the earth \(=\) \(6400 \mathrm{~km})\) is
270601
The radius and density of two artificial satellites are \(R_{1}, R_{2}\) and \(\rho_{1}, \rho_{2}\) respectively. The ratio of acceleration due to gravities on them will be
1 \(\frac{R_{2} \rho_{2}}{R_{1} \rho_{1}}\)
2 \(\frac{R_{1} \rho_{2}}{R_{2} \rho_{1}}\)
3 \(\frac{R_{1} \rho_{1}}{R_{2} \rho_{2}}\)
4 \(\frac{R_{2} \rho_{1}}{R_{1} \rho_{2}}\)
Explanation:
\(g=\frac{4}{3} \pi G R \rho \Rightarrow g \propto R \rho\)
Gravitation
270602
A man weigh ' \(W\) ' on the surface of earth and his weight at a height ' \(R\) ' from surface of earth is ( \(R\) is Radius of earth)
1 \(\frac{W}{4}\)
2 \(\frac{W}{2}\)
3 \(\mathrm{W}\)
4 \(4 \mathrm{~W}\)
Explanation:
\(W=m g ; W^{\prime}=m g^{\prime}=m g \because \frac{R}{R+h} \overbrace{}^{2}=\frac{W}{4}\)
Gravitation
270603
The acceleration due to gravity at the latitude \(45^{\circ}\) on the earth becomes zero if the angular velocity of rotation of earth is