NEWTON'S LAWS OF MOTION
Laws of Motion

270360 A lift is going up, the total mass of the lift and the passengers is\(1500 \mathrm{~kg}\). The variation in the speed of lift is shown in fig. Then the tension in the rope at \(t=1 \mathrm{~s}\) will be

\(\mu=0.6\)
\(\mu=0.5\)

1 \(17400 \mathrm{~N}\)
2 \(14700 \mathrm{~N}\)
3 \(12000 \mathrm{~N}\)
4 \(10000 \mathrm{~N}\)
Laws of Motion

270361 In the above problem the tension in the rope will be least at

1 \(t=1 \mathrm{~s}\)
2 \(t=4 \mathrm{~s}\)
3 \(t=9 s\)
4 \(\mathrm{t}=11 \mathrm{~s}\)
Laws of Motion

270362 A piece of wire is bent in the shape of a parabola \(y=k x^{2}\) ( \(y\)-axis vertical) with a bead of mass \(m\) on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the \(x\) axis with a constant acceleration \(a\). The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the \(y\)-axis is :

1 \(\frac{a}{g k}\)
2 \(\frac{a}{2 g k}\)
3 \(\frac{2 a}{g k}\)
4 \(\frac{a}{4 g k}\)
Laws of Motion

270360 A lift is going up, the total mass of the lift and the passengers is\(1500 \mathrm{~kg}\). The variation in the speed of lift is shown in fig. Then the tension in the rope at \(t=1 \mathrm{~s}\) will be

\(\mu=0.6\)
\(\mu=0.5\)

1 \(17400 \mathrm{~N}\)
2 \(14700 \mathrm{~N}\)
3 \(12000 \mathrm{~N}\)
4 \(10000 \mathrm{~N}\)
Laws of Motion

270361 In the above problem the tension in the rope will be least at

1 \(t=1 \mathrm{~s}\)
2 \(t=4 \mathrm{~s}\)
3 \(t=9 s\)
4 \(\mathrm{t}=11 \mathrm{~s}\)
Laws of Motion

270362 A piece of wire is bent in the shape of a parabola \(y=k x^{2}\) ( \(y\)-axis vertical) with a bead of mass \(m\) on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the \(x\) axis with a constant acceleration \(a\). The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the \(y\)-axis is :

1 \(\frac{a}{g k}\)
2 \(\frac{a}{2 g k}\)
3 \(\frac{2 a}{g k}\)
4 \(\frac{a}{4 g k}\)
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Laws of Motion

270360 A lift is going up, the total mass of the lift and the passengers is\(1500 \mathrm{~kg}\). The variation in the speed of lift is shown in fig. Then the tension in the rope at \(t=1 \mathrm{~s}\) will be

\(\mu=0.6\)
\(\mu=0.5\)

1 \(17400 \mathrm{~N}\)
2 \(14700 \mathrm{~N}\)
3 \(12000 \mathrm{~N}\)
4 \(10000 \mathrm{~N}\)
Laws of Motion

270361 In the above problem the tension in the rope will be least at

1 \(t=1 \mathrm{~s}\)
2 \(t=4 \mathrm{~s}\)
3 \(t=9 s\)
4 \(\mathrm{t}=11 \mathrm{~s}\)
Laws of Motion

270362 A piece of wire is bent in the shape of a parabola \(y=k x^{2}\) ( \(y\)-axis vertical) with a bead of mass \(m\) on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the \(x\) axis with a constant acceleration \(a\). The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the \(y\)-axis is :

1 \(\frac{a}{g k}\)
2 \(\frac{a}{2 g k}\)
3 \(\frac{2 a}{g k}\)
4 \(\frac{a}{4 g k}\)