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A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?
A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 ms−1 relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley. How much has the trolley moved from the time the child begins to run?
Explanation: Mass of troly M = 200Kg mass of child m = 20Kg speed of trolley v = 36Km/hr=36 x 5/18 = 10m/s Let v' be the final velocity of the trolley with respect to the ground. Final velocity of the boy with respect to the ground = v,−4 from conservation of linear momentum
If a man increases his speed by 2 m/sec, his K.E. is doubled. The original speed of the man is-
A force Newton (where x is in metres) acts on a particle which moves from a position (2m, 3m) to (3m, 0m). Then the work done is
The angular position of a particle (in radians), along a circle of radius 0.8 m is given by the function in time (seconds) by The linear velocity of the particle
The answer is option A
So, as we know,
After differentiating angular velocity with respect to t,
Linear velocity=2t+2.3=ω
Now,
Velocity=r x ω(where r is 0.8m)
=0.8x2.3
=1.84m/s
Which of the following is not a unit of angular displacement?
Angular displacement is measured in units of radians. 2 pi radians equals 360 degrees. The angular displacement is not a length (i.e. not measured in meters or feet), so an angular displacement is different than a linear displacement.
A boy is playing with a tire of radius 0.5m. He accelerates it from 5rpm to 25 rpm in 15 seconds. The linear acceleration of tire is
r=0.5m, t=15s n1=5rpm=5/60 rps n2=25rpm=25/60rps … ω1=2π(5) rad/s … ω2=2π(25) rad/s As, a=r∝ [∝=ω2- ω1/t] so, a=0.5[2π(n2-n1)/t] a=0.5x6.28x20/60x15 a=6.28x2/60x3 a=6.28/90 a=0.697 m/s2 a≈0.7 m/s2
In an equilateral triangle of length 6 cm , three masses m1= 40g , m2= 60g and m3= 60g are located at the vertices. The moment of inertia of the system about an axis along the altitude of the triangle passing through m1 is
Axis is along AD so Moment of inertia of first mass is zero and 2nd and 3rd masses are at distance of 6/2 from axis, so by using simply I=mr2 we get, I=60×36/4+60×36/4=1080 g cm2
For a system to be in equilibrium, the net torques acting on it must balance. This is true only if the torque are taken about
For a system to be in equilibrium torques on the system must be equal to zero
This is true only if the torques are taken about the centre of mass as the whole mass of the object seems to be situated at centre of mass.
The Mechanical advantage of a lever is given by-
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