270162
A mass of \(10 \mathrm{~kg}\) is suspended by a rope of length \(2.8 \mathrm{~m}\) from a ceiling. A force of \(98 \mathrm{~N}\) is applied at the midpoint of the rope as shown in figure. The angle which the rope makes with the vertical in equilibrium is
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
According to Lami's theorem \(\frac{F_{1}}{\sin \alpha}=\frac{F_{2}}{\sin \beta}\)
Laws of Motion
270163
A mass of \(M \mathrm{~kg}\) is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle \(45^{\circ}\) with the initial vertical direction is
1 \(M g\)
2 \(\frac{M g}{\sqrt{2}}\)
3 \(\operatorname{Mg}(\sqrt{2}+1)\)
4 \(\sqrt{2} \mathrm{Mg}\)
Explanation:
\(\operatorname{Tan} \theta=\frac{F}{m g}\)
Laws of Motion
270267
Two masses \(M_{1}\) and \(M_{2}\) connected by means of a string which is made to pass over light, smooth pulley are in equilibrium on a fixed smooth wedge as shown in figure. If\(\theta=60^{\circ}\) and \(\alpha=30^{\circ}\), then the ratio of \(M_{1}\) to \(M_{2}\) is
1 \(1: 2\)
2 \(2: \sqrt{3}\)
3 \(1: \sqrt{3}\)
4 \(\sqrt{3}: 1\)
Explanation:
\(m_{1} g \sin \theta=m_{2} g \sin \alpha\)
Laws of Motion
270268
If '\(O\) ' is at equilibrium then the values of the tension \(T_{1}\) and \(T_{2}\) respectively.
270269
A\(1 \mathrm{~N}\) pendulum bob is held at an angle \(\theta\) from the vertical by a \(2 \mathrm{~N}\) horizontal force \(F\) as shown in the figure. The tension in the string supporting the pendulum bob (in newton) is
(2011E )
1 \(\cos \theta\)
2 \(\frac{2}{\cos \theta}\)
3 \(\sqrt{5}\)
4 1
Explanation:
\(T \cos \theta=W, T \sin \theta=F, T=\sqrt{W^{2}+F^{2}}\)
270162
A mass of \(10 \mathrm{~kg}\) is suspended by a rope of length \(2.8 \mathrm{~m}\) from a ceiling. A force of \(98 \mathrm{~N}\) is applied at the midpoint of the rope as shown in figure. The angle which the rope makes with the vertical in equilibrium is
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
According to Lami's theorem \(\frac{F_{1}}{\sin \alpha}=\frac{F_{2}}{\sin \beta}\)
Laws of Motion
270163
A mass of \(M \mathrm{~kg}\) is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle \(45^{\circ}\) with the initial vertical direction is
1 \(M g\)
2 \(\frac{M g}{\sqrt{2}}\)
3 \(\operatorname{Mg}(\sqrt{2}+1)\)
4 \(\sqrt{2} \mathrm{Mg}\)
Explanation:
\(\operatorname{Tan} \theta=\frac{F}{m g}\)
Laws of Motion
270267
Two masses \(M_{1}\) and \(M_{2}\) connected by means of a string which is made to pass over light, smooth pulley are in equilibrium on a fixed smooth wedge as shown in figure. If\(\theta=60^{\circ}\) and \(\alpha=30^{\circ}\), then the ratio of \(M_{1}\) to \(M_{2}\) is
1 \(1: 2\)
2 \(2: \sqrt{3}\)
3 \(1: \sqrt{3}\)
4 \(\sqrt{3}: 1\)
Explanation:
\(m_{1} g \sin \theta=m_{2} g \sin \alpha\)
Laws of Motion
270268
If '\(O\) ' is at equilibrium then the values of the tension \(T_{1}\) and \(T_{2}\) respectively.
270269
A\(1 \mathrm{~N}\) pendulum bob is held at an angle \(\theta\) from the vertical by a \(2 \mathrm{~N}\) horizontal force \(F\) as shown in the figure. The tension in the string supporting the pendulum bob (in newton) is
(2011E )
1 \(\cos \theta\)
2 \(\frac{2}{\cos \theta}\)
3 \(\sqrt{5}\)
4 1
Explanation:
\(T \cos \theta=W, T \sin \theta=F, T=\sqrt{W^{2}+F^{2}}\)
270162
A mass of \(10 \mathrm{~kg}\) is suspended by a rope of length \(2.8 \mathrm{~m}\) from a ceiling. A force of \(98 \mathrm{~N}\) is applied at the midpoint of the rope as shown in figure. The angle which the rope makes with the vertical in equilibrium is
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
According to Lami's theorem \(\frac{F_{1}}{\sin \alpha}=\frac{F_{2}}{\sin \beta}\)
Laws of Motion
270163
A mass of \(M \mathrm{~kg}\) is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle \(45^{\circ}\) with the initial vertical direction is
1 \(M g\)
2 \(\frac{M g}{\sqrt{2}}\)
3 \(\operatorname{Mg}(\sqrt{2}+1)\)
4 \(\sqrt{2} \mathrm{Mg}\)
Explanation:
\(\operatorname{Tan} \theta=\frac{F}{m g}\)
Laws of Motion
270267
Two masses \(M_{1}\) and \(M_{2}\) connected by means of a string which is made to pass over light, smooth pulley are in equilibrium on a fixed smooth wedge as shown in figure. If\(\theta=60^{\circ}\) and \(\alpha=30^{\circ}\), then the ratio of \(M_{1}\) to \(M_{2}\) is
1 \(1: 2\)
2 \(2: \sqrt{3}\)
3 \(1: \sqrt{3}\)
4 \(\sqrt{3}: 1\)
Explanation:
\(m_{1} g \sin \theta=m_{2} g \sin \alpha\)
Laws of Motion
270268
If '\(O\) ' is at equilibrium then the values of the tension \(T_{1}\) and \(T_{2}\) respectively.
270269
A\(1 \mathrm{~N}\) pendulum bob is held at an angle \(\theta\) from the vertical by a \(2 \mathrm{~N}\) horizontal force \(F\) as shown in the figure. The tension in the string supporting the pendulum bob (in newton) is
(2011E )
1 \(\cos \theta\)
2 \(\frac{2}{\cos \theta}\)
3 \(\sqrt{5}\)
4 1
Explanation:
\(T \cos \theta=W, T \sin \theta=F, T=\sqrt{W^{2}+F^{2}}\)
270162
A mass of \(10 \mathrm{~kg}\) is suspended by a rope of length \(2.8 \mathrm{~m}\) from a ceiling. A force of \(98 \mathrm{~N}\) is applied at the midpoint of the rope as shown in figure. The angle which the rope makes with the vertical in equilibrium is
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
According to Lami's theorem \(\frac{F_{1}}{\sin \alpha}=\frac{F_{2}}{\sin \beta}\)
Laws of Motion
270163
A mass of \(M \mathrm{~kg}\) is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle \(45^{\circ}\) with the initial vertical direction is
1 \(M g\)
2 \(\frac{M g}{\sqrt{2}}\)
3 \(\operatorname{Mg}(\sqrt{2}+1)\)
4 \(\sqrt{2} \mathrm{Mg}\)
Explanation:
\(\operatorname{Tan} \theta=\frac{F}{m g}\)
Laws of Motion
270267
Two masses \(M_{1}\) and \(M_{2}\) connected by means of a string which is made to pass over light, smooth pulley are in equilibrium on a fixed smooth wedge as shown in figure. If\(\theta=60^{\circ}\) and \(\alpha=30^{\circ}\), then the ratio of \(M_{1}\) to \(M_{2}\) is
1 \(1: 2\)
2 \(2: \sqrt{3}\)
3 \(1: \sqrt{3}\)
4 \(\sqrt{3}: 1\)
Explanation:
\(m_{1} g \sin \theta=m_{2} g \sin \alpha\)
Laws of Motion
270268
If '\(O\) ' is at equilibrium then the values of the tension \(T_{1}\) and \(T_{2}\) respectively.
270269
A\(1 \mathrm{~N}\) pendulum bob is held at an angle \(\theta\) from the vertical by a \(2 \mathrm{~N}\) horizontal force \(F\) as shown in the figure. The tension in the string supporting the pendulum bob (in newton) is
(2011E )
1 \(\cos \theta\)
2 \(\frac{2}{\cos \theta}\)
3 \(\sqrt{5}\)
4 1
Explanation:
\(T \cos \theta=W, T \sin \theta=F, T=\sqrt{W^{2}+F^{2}}\)
270162
A mass of \(10 \mathrm{~kg}\) is suspended by a rope of length \(2.8 \mathrm{~m}\) from a ceiling. A force of \(98 \mathrm{~N}\) is applied at the midpoint of the rope as shown in figure. The angle which the rope makes with the vertical in equilibrium is
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
According to Lami's theorem \(\frac{F_{1}}{\sin \alpha}=\frac{F_{2}}{\sin \beta}\)
Laws of Motion
270163
A mass of \(M \mathrm{~kg}\) is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle \(45^{\circ}\) with the initial vertical direction is
1 \(M g\)
2 \(\frac{M g}{\sqrt{2}}\)
3 \(\operatorname{Mg}(\sqrt{2}+1)\)
4 \(\sqrt{2} \mathrm{Mg}\)
Explanation:
\(\operatorname{Tan} \theta=\frac{F}{m g}\)
Laws of Motion
270267
Two masses \(M_{1}\) and \(M_{2}\) connected by means of a string which is made to pass over light, smooth pulley are in equilibrium on a fixed smooth wedge as shown in figure. If\(\theta=60^{\circ}\) and \(\alpha=30^{\circ}\), then the ratio of \(M_{1}\) to \(M_{2}\) is
1 \(1: 2\)
2 \(2: \sqrt{3}\)
3 \(1: \sqrt{3}\)
4 \(\sqrt{3}: 1\)
Explanation:
\(m_{1} g \sin \theta=m_{2} g \sin \alpha\)
Laws of Motion
270268
If '\(O\) ' is at equilibrium then the values of the tension \(T_{1}\) and \(T_{2}\) respectively.
270269
A\(1 \mathrm{~N}\) pendulum bob is held at an angle \(\theta\) from the vertical by a \(2 \mathrm{~N}\) horizontal force \(F\) as shown in the figure. The tension in the string supporting the pendulum bob (in newton) is
(2011E )
1 \(\cos \theta\)
2 \(\frac{2}{\cos \theta}\)
3 \(\sqrt{5}\)
4 1
Explanation:
\(T \cos \theta=W, T \sin \theta=F, T=\sqrt{W^{2}+F^{2}}\)