A pulley system is attached to an elevator as shown in figure. The elevator starts to move up with an acceleration $a$ .

Match statements in column I with results in column II.
$\begin{matrix} Column I & Column II (a) & Acceleration of m_{1} in elevator frame & (p) & \frac{1}{2} (\frac{(m_{2} - m_{1}) g + 2 m_{2} a}{m_{1} + m_{2}}) t^{2} (b) & Acceleration of m_{1} in ground frame & (q) & \frac{1}{2} [\frac{m_{2} - m_{1}}{(m_{1} + m_{2})} (g + a)] t^{2} (c) & Distance covered of m_{1} in elevator frame & (r) & (\frac{m_{2} - m_{1}}{m_{1} + m_{2}}) g + (\frac{2 m_{2}}{m_{1} + m_{2}}) a (d) & Distance covered of m_{1} in ground frame & (s) & \frac{(m_{2} - m_{1})}{(m_{1} + m_{2})} (g + a) (t) & None of the above \end{matrix}$

a - s; b - r; c - q; d - p

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a - r; b - q; c - p; d - s

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a - s; b - t; c - q; d - t

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a - r; b - q; c - s; d - p

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Solution

The correct option is A a - s; b - r; c - q; d - p

Consider an observer in elevator frame,

From figure,
$T - m_{1} (a + g) = m_{1} a_{1}$ ... (1)
$m_{2} (a + g) - T = m_{2} a_{1}$ ... (2)
From (1) and (2),
$m_{2} (a + g) - m_{1} (a + g) = (m_{1} + m_{2}) a_{1}$
$∴ a_{1} = \frac{(m_{2} - m_{1}) (a + g)}{(m_{1} + m_{2})}$
Distance covered in elevator frame
$= \frac{1}{2} (\frac{(m_{2} - m_{1}) (a + g)}{(m_{1} + m_{2})}) \times t^{2}$

In ground frame,

$a_{1} + a = \frac{m_{2} a - m_{1} a + g m_{2} - g m_{1} + a m_{1} + m_{2} a}{(m_{1} + m_{2})} = (\frac{m_{2} - m_{1}}{m_{1} + m_{2}}) g + (\frac{2 m_{2}}{m_{1} + m_{2}}) a$

Distance covered in ground frame $= \frac{1}{2} (\frac{(m_{2} - m_{1}) g + 2 m_{2} a}{m_{1} + m_{2}}) t^{2}$

(a) → (s); (b) → (r); (c) → (q); (d) → (p)

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Q. A pulley system is attached to an elevator as shown in figure. The elevator starts to move up with an acceleration

a

Match statements in column I with results in column II.

\begin{matrix} Column I & Column II (a) & Acceleration of m_{1} in elevator frame & (p) & \frac{1}{2} (\frac{(m_{2} - m_{1}) g + 2 m_{2} a}{m_{1} + m_{2}}) t^{2} (b) & Acceleration of m_{1} in ground frame & (q) & \frac{1}{2} [\frac{m_{2} - m_{1}}{(m_{1} + m_{2})} (g + a)] t^{2} (c) & Distance covered of m_{1} in elevator frame & (r) & (\frac{m_{2} - m_{1}}{m_{1} + m_{2}}) g + (\frac{2 m_{2}}{m_{1} + m_{2}}) a (d) & Distance covered of m_{1} in ground frame & (s) & \frac{(m_{2} - m_{1})}{(m_{1} + m_{2})} (g + a) (t) & None of the above \end{matrix}

A block of mass m = 15 kg is suspended in an elevator with the help of three identical light elastic cords (spring constant k = 100 N/m each) attached vertically. Cord 1 is tied to the ceiling of the elevator and the other two cords 2 and 3 are tied to the elevator floor as shown in the figure. When the elevator is stationary the tension force in each of the lower cords is T = 7.5 N. Take g = 10 $m / s^{2}$ . If now the elevator starts moving with given four accelerations shown in column I and Column II gives the displacement of the block with respect to elevator when it is accelerating. Column III gives tension in the cord 1 when elevator is accelerating.