A bob of mass 10 kg is attached to wire 0.3 m long. Its breaking stress is 4.8 x107 N/m2. The area of cross-section of the wire is 10-6 m2. The maximum angular velocity with which it can be rotated in a horizontal circle is
8 rad/s
4 rad/s
2 rad/s
1 rad/s
A rubber cord 10 m long is suspended vertically. How much does it stretch under its own weight? (Density of rubber is 1500 kg/m3, Y = 5 × 108 N/m2, g = 10 m/s2)
15 × 10-4 m
7.5 × 10-4 m
12 × 10-4 m
25 × 10-4 m
A brass rod of cross-sectional area 1 cm2 and length 0.2 m is compressed lengthwise by a weight of 5 kg of Young's modulus of elasticity of brass is 1 x 1011 N/m2 and g = 10 m/s2, then increase in the energy of the rod will be
10-5 J
2.5 × 10-5 J
5 × 10-5 J
2.5 × 10-4 J
A wire fixed at the upper end stretches by length by applying a force F. The work done in stretching is
FΔl
2 FΔl
The Young's modulus of a wire is numerically equal to the stress which will
not change the length of the wire
double the length of the wire
increase the length
change the radius of the wire to half
A one-metre long steel wire of cross-sectional area 1mm2 is extended by 1 mm. If Y=2 x 1011 N/m2, then the work done is :
0.1 J
0.2 J
0.3 J
0.4 J
A 5m aluminium wire (Y = 7 × 1010 N/m2) of diameter 3mm supports a 40 kg mass. In order to have the same elongation in copper wire (Y =12 x 1010 N/m2) of the same length under the same weight, the diameter should be in mm :
1.75
2.0
2.3
5.0
Minimum and maximum values of Poisson's ratio for a metal lies between:
- ∞ to + ∞
0 to 1
- ∞ to 1
0 to 0.5
A wire of diameter 1 mm breaks under a tension of 1000 N. Another wire, of same material as that of the first one, but of diameter 2 mm, breaks under a tension of :
500 N
100 N
1000 N
4000 N
A steel scale measures the length of a copper wire as 80.0 cm when both are at 20°C, the calibration temperature for the scale. What would the scale read for the length of the wire when both are at 40°C?
Given : α for steel = 11 × 10-6 per °C and for Cu= 17 × 10-6 per °C :
80.0096 cm
80.0272
1 cm
25.2 cm
A.
80.0096 cm
Using the relation
Increase in length of scale
lt = l0(1 + αt)
= 1 × [1 + 11×10-6 ×(40 - 20)]
= 1.00022 cm
Now, length of copper rod at 40°C
l't = l'0(1 + α't)
= 80[1 + 17×10-6 ×(40 - 20)]
= 80.0272 cm
Now, number of cms observed on the scale
= 80.0096