A 49 kg lady stands on a spring scale in an elevator. During the first 5 sec, starting from rest, the scale reads 69 kg. The velocity of the elevator will be
A. 10 m/sec
B. 15 m/sec
C. 20 m/sec
D. 25 m/sec
Answer: Option C
This Question Belongs to Civil Engineering >> Applied Mechanics And Graphic Statics
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In case of S.H.M. the period of oscillation (T), is given by
A. $${\text{T}} = \frac{{2\omega }}{{{\pi ^2}}}$$
B. $${\text{T}} = \frac{{2\pi }}{\omega }$$
C. $${\text{T}} = \frac{2}{\omega }$$
D. $${\text{T}} = \frac{\pi }{{2\omega }}$$
The angular speed of a car taking a circular turn of radius 100 m at 36 km/hr will be
A. 0.1 rad/sec
B. 1 rad/sec
C. 10 rad/sec
D. 100 rad/sec
A body is said to move with Simple Harmonic Motion if its acceleration, is
A. Always directed away from the centre, the point of reference
B. Proportional to the square of the distance from the point of reference
C. Proportional to the distance from the point of reference and directed towards it
D. Inversely proportion to the distance from the point of reference
The resultant of two forces P and Q acting at an angle $$\theta $$, is
A. $${{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{P}}\sin \theta $$
B. $${{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\cos \theta $$
C. $${{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\tan \theta $$
D. $$\sqrt {{{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\cos \theta } $$
E. $$\sqrt {{{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\sin \theta } $$
initial weight before moving on lift(m) =49 kg.
We know as we move upwards increase in weight occurs i.e acceleration increase by (g+a).
New weight (m1) = 69 kg.
Time(t) =5 sec.
Weight(W)= m(g+a) =m1 * g.
49(9.81+a) =69*9.81.
a=4m/sec2.
We know v=u+at.
u =0(as starts from rest).
v = 4*5=20 m /sec2.
the correct answer is [M-1 L3 T-2]
Asa ka kyo heee? Kyo ke +5 to app ne keya hee nai hiiiii