270494
The gravitational force between two bodies is \(6.67 \times 10^{-7} \mathrm{~N}\) when the distance between their centres is \(10 \mathrm{~m}\). If the mass of first body is 800 \(\mathrm{kg}\), then the mass of second body is
270495
Two identical spheres each of radius \(R\) are placed with their centres at a distance \(n R\), where \(n\) is integer greater than 2 . The gravitational force between them will be proportional to
1 \(1 / R^{4}\)
2 \(1 / R^{2}\)
3 \(R^{2}\)
4 \(R^{4}\)
Explanation:
\(F=\frac{G m_{1} m_{2}}{R^{2}} ;\) Here \(m=\frac{4}{3} \pi R^{3}\)
Gravitation
270496
A satellite is orbiting round the earth. If both gravitational force and centripetal force on the satellite is \(F\), then, net force acting on the satellite to revolve round the earth is
1 \(F / 2\)
2 F
3 \(2 \mathrm{~F}\)
4 Zero
Explanation:
Gravitational force provides centripetal force.
Gravitation
270497
Mass \(M=1\) unit is divided into two parts \(X\) and \((1-X)\). For a given separation the value of \(\mathrm{X}\) for which the gravitational force between them becomes maximum is
1 \(1 / 2\)
2 \(3 / 5\)
3 1
4 2
Explanation:
\(F=\frac{G \times m(1-x) m x}{R^{2}}\) is maximum when \(x=\frac{1}{2}\)
270494
The gravitational force between two bodies is \(6.67 \times 10^{-7} \mathrm{~N}\) when the distance between their centres is \(10 \mathrm{~m}\). If the mass of first body is 800 \(\mathrm{kg}\), then the mass of second body is
270495
Two identical spheres each of radius \(R\) are placed with their centres at a distance \(n R\), where \(n\) is integer greater than 2 . The gravitational force between them will be proportional to
1 \(1 / R^{4}\)
2 \(1 / R^{2}\)
3 \(R^{2}\)
4 \(R^{4}\)
Explanation:
\(F=\frac{G m_{1} m_{2}}{R^{2}} ;\) Here \(m=\frac{4}{3} \pi R^{3}\)
Gravitation
270496
A satellite is orbiting round the earth. If both gravitational force and centripetal force on the satellite is \(F\), then, net force acting on the satellite to revolve round the earth is
1 \(F / 2\)
2 F
3 \(2 \mathrm{~F}\)
4 Zero
Explanation:
Gravitational force provides centripetal force.
Gravitation
270497
Mass \(M=1\) unit is divided into two parts \(X\) and \((1-X)\). For a given separation the value of \(\mathrm{X}\) for which the gravitational force between them becomes maximum is
1 \(1 / 2\)
2 \(3 / 5\)
3 1
4 2
Explanation:
\(F=\frac{G \times m(1-x) m x}{R^{2}}\) is maximum when \(x=\frac{1}{2}\)
270494
The gravitational force between two bodies is \(6.67 \times 10^{-7} \mathrm{~N}\) when the distance between their centres is \(10 \mathrm{~m}\). If the mass of first body is 800 \(\mathrm{kg}\), then the mass of second body is
270495
Two identical spheres each of radius \(R\) are placed with their centres at a distance \(n R\), where \(n\) is integer greater than 2 . The gravitational force between them will be proportional to
1 \(1 / R^{4}\)
2 \(1 / R^{2}\)
3 \(R^{2}\)
4 \(R^{4}\)
Explanation:
\(F=\frac{G m_{1} m_{2}}{R^{2}} ;\) Here \(m=\frac{4}{3} \pi R^{3}\)
Gravitation
270496
A satellite is orbiting round the earth. If both gravitational force and centripetal force on the satellite is \(F\), then, net force acting on the satellite to revolve round the earth is
1 \(F / 2\)
2 F
3 \(2 \mathrm{~F}\)
4 Zero
Explanation:
Gravitational force provides centripetal force.
Gravitation
270497
Mass \(M=1\) unit is divided into two parts \(X\) and \((1-X)\). For a given separation the value of \(\mathrm{X}\) for which the gravitational force between them becomes maximum is
1 \(1 / 2\)
2 \(3 / 5\)
3 1
4 2
Explanation:
\(F=\frac{G \times m(1-x) m x}{R^{2}}\) is maximum when \(x=\frac{1}{2}\)
270494
The gravitational force between two bodies is \(6.67 \times 10^{-7} \mathrm{~N}\) when the distance between their centres is \(10 \mathrm{~m}\). If the mass of first body is 800 \(\mathrm{kg}\), then the mass of second body is
270495
Two identical spheres each of radius \(R\) are placed with their centres at a distance \(n R\), where \(n\) is integer greater than 2 . The gravitational force between them will be proportional to
1 \(1 / R^{4}\)
2 \(1 / R^{2}\)
3 \(R^{2}\)
4 \(R^{4}\)
Explanation:
\(F=\frac{G m_{1} m_{2}}{R^{2}} ;\) Here \(m=\frac{4}{3} \pi R^{3}\)
Gravitation
270496
A satellite is orbiting round the earth. If both gravitational force and centripetal force on the satellite is \(F\), then, net force acting on the satellite to revolve round the earth is
1 \(F / 2\)
2 F
3 \(2 \mathrm{~F}\)
4 Zero
Explanation:
Gravitational force provides centripetal force.
Gravitation
270497
Mass \(M=1\) unit is divided into two parts \(X\) and \((1-X)\). For a given separation the value of \(\mathrm{X}\) for which the gravitational force between them becomes maximum is
1 \(1 / 2\)
2 \(3 / 5\)
3 1
4 2
Explanation:
\(F=\frac{G \times m(1-x) m x}{R^{2}}\) is maximum when \(x=\frac{1}{2}\)
