(1)The breaking strength of a rope is one and a half times the weight of a stone. The minimum time in which this stone can be raised through 10m using this rope is nearly
(a) 0.5s (b)1s (c) 2s (d) 2.5s (e) 3s
The minimum time will be obtained when the acceleration with which the stone is pulled up is maximum. But the maximum possible acceleration ‘a’ is that which will make the apparent weight of the stone(as in the case of a body in a lift) one and a half times its real weight. Therefore we have, m(g+a) = 1.5mg from which a=0.5g.
Substituting this value of ‘a’ in the equation, s = ut + ½ at^2, we have
10 = ½ ×0.5g t^2 since u=0. Taking ‘g’ to be nearly 10m/s^2 we obtain t = 2s.
(2) A rocket with a lift off mass 3.5×10^4 kg is blasted upwards with an initial acceleration of 10m/s^2. Then the initial thrust of the blast is
(a) 1.75×10^5N (b) 3.5×10^5N (c) 7×10^5N (d) 14×10^5N
The initial thrust of the blast is equal in magnitude to the apparent weight of the rocket while moving up with the given acceleration(as in the case of a body in a lift).Therefore, thrust = m(g+a) = 3.5×10^4(10+10) = 7×10^5N [Option (c)]This question appeared in the A.I.E.E.Exmination question paper of 2003.
(a) 0.5s (b)1s (c) 2s (d) 2.5s (e) 3s
The minimum time will be obtained when the acceleration with which the stone is pulled up is maximum. But the maximum possible acceleration ‘a’ is that which will make the apparent weight of the stone(as in the case of a body in a lift) one and a half times its real weight. Therefore we have, m(g+a) = 1.5mg from which a=0.5g.
Substituting this value of ‘a’ in the equation, s = ut + ½ at^2, we have
10 = ½ ×0.5g t^2 since u=0. Taking ‘g’ to be nearly 10m/s^2 we obtain t = 2s.
(2) A rocket with a lift off mass 3.5×10^4 kg is blasted upwards with an initial acceleration of 10m/s^2. Then the initial thrust of the blast is
(a) 1.75×10^5N (b) 3.5×10^5N (c) 7×10^5N (d) 14×10^5N
The initial thrust of the blast is equal in magnitude to the apparent weight of the rocket while moving up with the given acceleration(as in the case of a body in a lift).Therefore, thrust = m(g+a) = 3.5×10^4(10+10) = 7×10^5N [Option (c)]This question appeared in the A.I.E.E.Exmination question paper of 2003.
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