A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is: (take g = 10 ms−2 )
A uniform string of length 20 m is suspended from a rigid support. Such that the time taken to reach the support.
A satellite is revolving in a circular orbit at a height ‘h’ from the earth’s surface (radius of earth R ; h<<R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth’s gravitational field, is close to: (Neglect the effect of atmosphere.)
Given, a satellite is revolving in a circular orbit at a height h from the Earth's surface having radius of earth R, i.e h <<R
Orbit velocity of a satellite
therefore, the minimum increase in its orbital velocity required to escape from the Earth's Gravitational Field.
A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?
Given that the screw gauge has zero error.
So, least count of screw gauge = 0.5/50 mm
The thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line. we have
=0 .50 mm + (25) x 0.5/50 mm
= 0.50 mm + 0.25 mm = 0.75 mm
A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g=9.8 ms−2:
2.45 ×10−3 kg
6.45 x×10−3 kg
9.89 ×10−3 kg
12.89 ×10−3 kg
12.89 ×10−3 kg
Given potential energy burnt by lifting weight
= mgh = 10 x 9.8 x 1 x 1000 = 9.8 x 104 J
If mass lost by a person be m, then energy dissipated
= m x 2 x 38 x 107 J /10
⇒ m = 5 x 10-3 x 9.8 / 3.8
= 12.89 x 10-3 kg
A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be:
Arithmetic mean time of an oscillating simple pendulum
Mean deviation of a simple pendulum
Given, minimum division in the measuring clock, i.e. simple pendulum = 1s. Thus, the reported mean time of an oscillating simple pendulum =
A particle of mass m is moving along the side of a square of side ‘a’, with a uniform speed v in the x-y plane as shown in the figure:
Which of the following statements is false for the angular momentum → L about the origin?
For a particle of mass, m is moving along the side of a square a. Such that
Angular momentum L about the origin
A roller is made by joining together two cones at their vertices O. It is kept on two rails AB and CD which are placed asymmetrically (see figure), with its axis perpendicular to CD and its centre O at the centre of line joining AB and CD (see figure). It is given a light push so that it starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to:
turn left and right alternately
As, the wheel rolls forward the radius of the wheel, decreases along AB hence for the same number of rotations it moves less distance along AB, hence it turns left.
A pendulum clock loses 12 s a day if the temperature is 408C and gains 4 s a day if the temperature is 208C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion (α) of the metal of the pendulum shaft are respectively:
25 C; α=1.85×10−5/ °C
60 °C; α=1.85×10−4/ °C
25 C; α=1.85×10−5/ °C
Time period of pendulum,
Thus, the coefficient of linear expansion in pendulum clock = 1.85 x 10-5/C
A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals µ. The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction µ and the distance x(=QR), are, respectively close to :
0.2 and 6.5 m
0.2 and 3.5 m
0.29 and 3.5 m
0.29 and 6.5 m
0.29 and 3.5 m
Energy lost over path PQ = μ mg cos θ x 4
Energy lost over path QR = μ mgx
i.e μ mg cos 30° x 4 = μ mgx (∴ θ = 30°)
From Q to R energy loss is half of the total energy loss.
i.e μ mgx = mgh/2
μ = 0.29
The values of the coefficient of friction μ and the distance x (=OR) are 0.29 and 3.5 m
An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PVn=constant, then n is given by (Here CP and CV are molar specific heat at constant pressure and constant volume, respectively):
For the polytropic process, specific heat for an ideal gas.