270546
A \(3 \mathrm{~kg}\) mass and a \(4 \mathrm{~kg}\) mass are placed on \(x\) and \(y\) axes at a distance of 1 metre from the origin and a \(1 \mathrm{~kg}\) mass is placed at the origin. Then the resultant gravitational force on \(1 \mathrm{~kg}\) mass is
270592
Gravitational force between two point masses \(m\) and \(M\) separated by a distance \(r\) is \(F\). Now if a point mass \(3 \mathrm{~m}\) is placed next to \(\mathrm{m}\), the total force on \(M\) will be
1 \(\mathrm{F}\)
2 \(2 \mathrm{~F}\)
3 \(3 \mathrm{~F}\)
4 \(4 \mathrm{~F}\)
Explanation:
\(F=\frac{G M m}{r^{2}} ; F^{\prime}=\frac{G M(m+2 m)}{r^{2}}\)
Gravitation
270593
If three particles,each of mass \(M\) are placed at the three corners of an equilateral triangle of side \(a\), the force exerted by this system on another particle of mass \(M\) placed (i) at the midpoint of a side and (ii) at the centre of the triangle are, respectively.
1 \(0, \frac{4 G M^{2}}{3 a^{2}}\)
2 \(\frac{4 G M^{2}}{3 a^{2}}, 0\)
3 \(\frac{3 G M^{2}}{a^{2}}, \frac{G M^{2}}{a^{2}}\)
4 0,0
Explanation:
Find individual forces and calculate resultant
\(\text { Use } F=\frac{G m_{1} m_{2}}{R^{2}}\)
Gravitation
270594
Two masses ' \(M\) ' and ' \(4 M\) ' are at a distance ' \(r\) ' apart on the line joining them, ' \(P\) ' is point where the resultant gravitational force is zero (such a point is called as null point). The distance of ' \(P\) ' from the mass ' \(M\) ' is
270546
A \(3 \mathrm{~kg}\) mass and a \(4 \mathrm{~kg}\) mass are placed on \(x\) and \(y\) axes at a distance of 1 metre from the origin and a \(1 \mathrm{~kg}\) mass is placed at the origin. Then the resultant gravitational force on \(1 \mathrm{~kg}\) mass is
270592
Gravitational force between two point masses \(m\) and \(M\) separated by a distance \(r\) is \(F\). Now if a point mass \(3 \mathrm{~m}\) is placed next to \(\mathrm{m}\), the total force on \(M\) will be
1 \(\mathrm{F}\)
2 \(2 \mathrm{~F}\)
3 \(3 \mathrm{~F}\)
4 \(4 \mathrm{~F}\)
Explanation:
\(F=\frac{G M m}{r^{2}} ; F^{\prime}=\frac{G M(m+2 m)}{r^{2}}\)
Gravitation
270593
If three particles,each of mass \(M\) are placed at the three corners of an equilateral triangle of side \(a\), the force exerted by this system on another particle of mass \(M\) placed (i) at the midpoint of a side and (ii) at the centre of the triangle are, respectively.
1 \(0, \frac{4 G M^{2}}{3 a^{2}}\)
2 \(\frac{4 G M^{2}}{3 a^{2}}, 0\)
3 \(\frac{3 G M^{2}}{a^{2}}, \frac{G M^{2}}{a^{2}}\)
4 0,0
Explanation:
Find individual forces and calculate resultant
\(\text { Use } F=\frac{G m_{1} m_{2}}{R^{2}}\)
Gravitation
270594
Two masses ' \(M\) ' and ' \(4 M\) ' are at a distance ' \(r\) ' apart on the line joining them, ' \(P\) ' is point where the resultant gravitational force is zero (such a point is called as null point). The distance of ' \(P\) ' from the mass ' \(M\) ' is
270546
A \(3 \mathrm{~kg}\) mass and a \(4 \mathrm{~kg}\) mass are placed on \(x\) and \(y\) axes at a distance of 1 metre from the origin and a \(1 \mathrm{~kg}\) mass is placed at the origin. Then the resultant gravitational force on \(1 \mathrm{~kg}\) mass is
270592
Gravitational force between two point masses \(m\) and \(M\) separated by a distance \(r\) is \(F\). Now if a point mass \(3 \mathrm{~m}\) is placed next to \(\mathrm{m}\), the total force on \(M\) will be
1 \(\mathrm{F}\)
2 \(2 \mathrm{~F}\)
3 \(3 \mathrm{~F}\)
4 \(4 \mathrm{~F}\)
Explanation:
\(F=\frac{G M m}{r^{2}} ; F^{\prime}=\frac{G M(m+2 m)}{r^{2}}\)
Gravitation
270593
If three particles,each of mass \(M\) are placed at the three corners of an equilateral triangle of side \(a\), the force exerted by this system on another particle of mass \(M\) placed (i) at the midpoint of a side and (ii) at the centre of the triangle are, respectively.
1 \(0, \frac{4 G M^{2}}{3 a^{2}}\)
2 \(\frac{4 G M^{2}}{3 a^{2}}, 0\)
3 \(\frac{3 G M^{2}}{a^{2}}, \frac{G M^{2}}{a^{2}}\)
4 0,0
Explanation:
Find individual forces and calculate resultant
\(\text { Use } F=\frac{G m_{1} m_{2}}{R^{2}}\)
Gravitation
270594
Two masses ' \(M\) ' and ' \(4 M\) ' are at a distance ' \(r\) ' apart on the line joining them, ' \(P\) ' is point where the resultant gravitational force is zero (such a point is called as null point). The distance of ' \(P\) ' from the mass ' \(M\) ' is
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Gravitation
270546
A \(3 \mathrm{~kg}\) mass and a \(4 \mathrm{~kg}\) mass are placed on \(x\) and \(y\) axes at a distance of 1 metre from the origin and a \(1 \mathrm{~kg}\) mass is placed at the origin. Then the resultant gravitational force on \(1 \mathrm{~kg}\) mass is
270592
Gravitational force between two point masses \(m\) and \(M\) separated by a distance \(r\) is \(F\). Now if a point mass \(3 \mathrm{~m}\) is placed next to \(\mathrm{m}\), the total force on \(M\) will be
1 \(\mathrm{F}\)
2 \(2 \mathrm{~F}\)
3 \(3 \mathrm{~F}\)
4 \(4 \mathrm{~F}\)
Explanation:
\(F=\frac{G M m}{r^{2}} ; F^{\prime}=\frac{G M(m+2 m)}{r^{2}}\)
Gravitation
270593
If three particles,each of mass \(M\) are placed at the three corners of an equilateral triangle of side \(a\), the force exerted by this system on another particle of mass \(M\) placed (i) at the midpoint of a side and (ii) at the centre of the triangle are, respectively.
1 \(0, \frac{4 G M^{2}}{3 a^{2}}\)
2 \(\frac{4 G M^{2}}{3 a^{2}}, 0\)
3 \(\frac{3 G M^{2}}{a^{2}}, \frac{G M^{2}}{a^{2}}\)
4 0,0
Explanation:
Find individual forces and calculate resultant
\(\text { Use } F=\frac{G m_{1} m_{2}}{R^{2}}\)
Gravitation
270594
Two masses ' \(M\) ' and ' \(4 M\) ' are at a distance ' \(r\) ' apart on the line joining them, ' \(P\) ' is point where the resultant gravitational force is zero (such a point is called as null point). The distance of ' \(P\) ' from the mass ' \(M\) ' is