06. Motion of Body Connected Together
Laws of Motion

146389 Consider the following statement about the blocks, shown in the diagram, which are being pushed by a constant force on a frictionless table.

1 A only
2 B only
3 Both A and B
4 neither A nor B
Laws of Motion

146391 The weight of a man in a lift, moving upwards is \(608 \mathrm{~N}\) while the weight of the same man in the lift moving downwards, with the same acceleration, is \(368 \mathrm{~N}\). His normal weight in newton is:

1 488
2 588
3 480
4 240
Laws of Motion

146392 In the system shown in the adjoining figure, the tension \(T_{2}\) is

1 \(g\)
2 \(2 \mathrm{~g}\)
3 \(5 \mathrm{~g}\)
4 \(6 \mathrm{~g}\)
Laws of Motion

146393 Four blocks of same mass connected by cords are pulled by a force \(F\) on a smooth horizontal surface, as shown in figure. The tensions \(T_{1}, T_{2}\) and \(T_{3}\) will be
\(\mathrm{F} \longleftarrow \mathrm{M} \stackrel{\mathrm{T}_{1}}{\longleftarrow} \mathrm{M} \stackrel{\mathrm{T}_{2}}{\longleftarrow} \mathrm{M} \stackrel{\mathrm{T}_{3}}{\longleftarrow} \mathrm{M}\)

1 \(\mathrm{T}_{1}=\frac{1}{4} \mathrm{~F}, \mathrm{~T}_{2}=\frac{3}{2} \mathrm{~F}, \mathrm{~T}_{3}=\frac{1}{4} \mathrm{~F}\)
2 \(\mathrm{T}_{1}=\frac{1}{4} \mathrm{~F}, \mathrm{~T}_{2}=\frac{1}{2} \mathrm{~F}, \mathrm{~T}_{3}=\frac{1}{2} \mathrm{~F}\)
3 \(\mathrm{T}_{1}=\frac{3}{4} \mathrm{~F}, \mathrm{~T}_{2}=\frac{1}{2} \mathrm{~F}, \mathrm{~T}_{3}=\frac{1}{4} \mathrm{~F}\)
4 \(\mathrm{T}_{1}=\frac{3}{4} \mathrm{~F}, \mathrm{~T}_{2}=\frac{1}{2} \mathrm{~F}, \mathrm{~T}_{3}=\frac{1}{2} \mathrm{~F}\)
Laws of Motion

146389 Consider the following statement about the blocks, shown in the diagram, which are being pushed by a constant force on a frictionless table.

1 A only
2 B only
3 Both A and B
4 neither A nor B
Laws of Motion

146391 The weight of a man in a lift, moving upwards is \(608 \mathrm{~N}\) while the weight of the same man in the lift moving downwards, with the same acceleration, is \(368 \mathrm{~N}\). His normal weight in newton is:

1 488
2 588
3 480
4 240
Laws of Motion

146392 In the system shown in the adjoining figure, the tension \(T_{2}\) is

1 \(g\)
2 \(2 \mathrm{~g}\)
3 \(5 \mathrm{~g}\)
4 \(6 \mathrm{~g}\)
Laws of Motion

146393 Four blocks of same mass connected by cords are pulled by a force \(F\) on a smooth horizontal surface, as shown in figure. The tensions \(T_{1}, T_{2}\) and \(T_{3}\) will be
\(\mathrm{F} \longleftarrow \mathrm{M} \stackrel{\mathrm{T}_{1}}{\longleftarrow} \mathrm{M} \stackrel{\mathrm{T}_{2}}{\longleftarrow} \mathrm{M} \stackrel{\mathrm{T}_{3}}{\longleftarrow} \mathrm{M}\)

1 \(\mathrm{T}_{1}=\frac{1}{4} \mathrm{~F}, \mathrm{~T}_{2}=\frac{3}{2} \mathrm{~F}, \mathrm{~T}_{3}=\frac{1}{4} \mathrm{~F}\)
2 \(\mathrm{T}_{1}=\frac{1}{4} \mathrm{~F}, \mathrm{~T}_{2}=\frac{1}{2} \mathrm{~F}, \mathrm{~T}_{3}=\frac{1}{2} \mathrm{~F}\)
3 \(\mathrm{T}_{1}=\frac{3}{4} \mathrm{~F}, \mathrm{~T}_{2}=\frac{1}{2} \mathrm{~F}, \mathrm{~T}_{3}=\frac{1}{4} \mathrm{~F}\)
4 \(\mathrm{T}_{1}=\frac{3}{4} \mathrm{~F}, \mathrm{~T}_{2}=\frac{1}{2} \mathrm{~F}, \mathrm{~T}_{3}=\frac{1}{2} \mathrm{~F}\)
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Laws of Motion

146389 Consider the following statement about the blocks, shown in the diagram, which are being pushed by a constant force on a frictionless table.

1 A only
2 B only
3 Both A and B
4 neither A nor B
Laws of Motion

146391 The weight of a man in a lift, moving upwards is \(608 \mathrm{~N}\) while the weight of the same man in the lift moving downwards, with the same acceleration, is \(368 \mathrm{~N}\). His normal weight in newton is:

1 488
2 588
3 480
4 240
Laws of Motion

146392 In the system shown in the adjoining figure, the tension \(T_{2}\) is

1 \(g\)
2 \(2 \mathrm{~g}\)
3 \(5 \mathrm{~g}\)
4 \(6 \mathrm{~g}\)
Laws of Motion

146393 Four blocks of same mass connected by cords are pulled by a force \(F\) on a smooth horizontal surface, as shown in figure. The tensions \(T_{1}, T_{2}\) and \(T_{3}\) will be
\(\mathrm{F} \longleftarrow \mathrm{M} \stackrel{\mathrm{T}_{1}}{\longleftarrow} \mathrm{M} \stackrel{\mathrm{T}_{2}}{\longleftarrow} \mathrm{M} \stackrel{\mathrm{T}_{3}}{\longleftarrow} \mathrm{M}\)

1 \(\mathrm{T}_{1}=\frac{1}{4} \mathrm{~F}, \mathrm{~T}_{2}=\frac{3}{2} \mathrm{~F}, \mathrm{~T}_{3}=\frac{1}{4} \mathrm{~F}\)
2 \(\mathrm{T}_{1}=\frac{1}{4} \mathrm{~F}, \mathrm{~T}_{2}=\frac{1}{2} \mathrm{~F}, \mathrm{~T}_{3}=\frac{1}{2} \mathrm{~F}\)
3 \(\mathrm{T}_{1}=\frac{3}{4} \mathrm{~F}, \mathrm{~T}_{2}=\frac{1}{2} \mathrm{~F}, \mathrm{~T}_{3}=\frac{1}{4} \mathrm{~F}\)
4 \(\mathrm{T}_{1}=\frac{3}{4} \mathrm{~F}, \mathrm{~T}_{2}=\frac{1}{2} \mathrm{~F}, \mathrm{~T}_{3}=\frac{1}{2} \mathrm{~F}\)
Laws of Motion

146389 Consider the following statement about the blocks, shown in the diagram, which are being pushed by a constant force on a frictionless table.

1 A only
2 B only
3 Both A and B
4 neither A nor B
Laws of Motion

146391 The weight of a man in a lift, moving upwards is \(608 \mathrm{~N}\) while the weight of the same man in the lift moving downwards, with the same acceleration, is \(368 \mathrm{~N}\). His normal weight in newton is:

1 488
2 588
3 480
4 240
Laws of Motion

146392 In the system shown in the adjoining figure, the tension \(T_{2}\) is

1 \(g\)
2 \(2 \mathrm{~g}\)
3 \(5 \mathrm{~g}\)
4 \(6 \mathrm{~g}\)
Laws of Motion

146393 Four blocks of same mass connected by cords are pulled by a force \(F\) on a smooth horizontal surface, as shown in figure. The tensions \(T_{1}, T_{2}\) and \(T_{3}\) will be
\(\mathrm{F} \longleftarrow \mathrm{M} \stackrel{\mathrm{T}_{1}}{\longleftarrow} \mathrm{M} \stackrel{\mathrm{T}_{2}}{\longleftarrow} \mathrm{M} \stackrel{\mathrm{T}_{3}}{\longleftarrow} \mathrm{M}\)

1 \(\mathrm{T}_{1}=\frac{1}{4} \mathrm{~F}, \mathrm{~T}_{2}=\frac{3}{2} \mathrm{~F}, \mathrm{~T}_{3}=\frac{1}{4} \mathrm{~F}\)
2 \(\mathrm{T}_{1}=\frac{1}{4} \mathrm{~F}, \mathrm{~T}_{2}=\frac{1}{2} \mathrm{~F}, \mathrm{~T}_{3}=\frac{1}{2} \mathrm{~F}\)
3 \(\mathrm{T}_{1}=\frac{3}{4} \mathrm{~F}, \mathrm{~T}_{2}=\frac{1}{2} \mathrm{~F}, \mathrm{~T}_{3}=\frac{1}{4} \mathrm{~F}\)
4 \(\mathrm{T}_{1}=\frac{3}{4} \mathrm{~F}, \mathrm{~T}_{2}=\frac{1}{2} \mathrm{~F}, \mathrm{~T}_{3}=\frac{1}{2} \mathrm{~F}\)