Let W and W′ be the respective weights of the metre stick and the coin.
The mass of the metre stick is concentrated at its mid - point, i.e., at the 50 cm mark.
Mass of the meter stick = m’
Mass of each coin, m = 5g
When the coins are placed 12 cm away from the end P, the centre of mass gets shifted by 5 cm from point R toward the end P. The centre of mass is located at a distance of 45 cm from point P.
The net torque will be conserved for rotational equilibrium about point R.
10 × g (45 - 12) - m' g (50 - 45) = 0
∴ m' = \(\frac{10 × 33 }{ 5}\)= 66 g
Hence, the mass of the metre stick is 66 g.
The oxygen molecule has a mass of 5.30 × 10-26 kg and a moment of inertia of 1.94 ×10-46 kg m2 about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.
From a uniform disk of radius R, a circular hole of radius \(\frac{R}{2}\) is cut out. The centre of the hole is at \(\frac{R}{2}\) from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
To maintain a rotor at a uniform angular speed of 200 rad s-1, an engine needs to transmit a torque of 180 N m. What is the power required by the engine ? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.
Give reasons for the following.
(i) King Tut’s body has been subjected to repeated scrutiny.
(ii) Howard Carter’s investigation was resented.
(iii) Carter had to chisel away the solidified resins to raise the king’s remains.
(iv) Tut’s body was buried along with gilded treasures.
(v) The boy king changed his name from Tutankhaten to Tutankhamun.
Find the mean deviation about the median for the data
xi | 15 | 21 | 27 | 30 | 35 |
fi | 3 | 5 | 6 | 7 | 8 |