Steel and copper wires of the same length and area are stretched by the same weight one after the other. Young's modulus of steel and copper are \(2\times10^{11} ~\text{N/m}^2\) and \(1.2\times10^{11}~\text{N/m}^2\). The ratio of increase in length is:
1. | \(2 \over 5\) | 2. | \(3 \over 5\) |
3. | \(5 \over 4\) | 4. | \(5 \over 2\) |
1. | \({AE} \frac{R}{r} \) | 2. | \(A E \left(\frac{R-r}{r}\right)\) |
3. | \(\frac{E}{A}\left(\frac{R-r}{A}\right)\) | 4. | \(\frac{Er}{AR}\) |
If the ratio of lengths, radii and Young's modulus of steel and brass wires in the figure are \(a,\) \(b\) and \(c\) respectively, then the corresponding ratio of increase in their lengths will be:
1. | 2. | ||
3. | 4. |
Two wires are made of the same material and have the same volume. The first wire has a cross-sectional area \(A\) and the second wire has a cross-sectional area \(3A\). If the length of the first wire is increased by \(\Delta l\) on applying a force \(F\), how much force is needed to stretch the second wire by the same amount?
1. | \(9F\) | 2. | \(6F\) |
3. | \(4F\) | 4. | \(F\) |
1. | \(1:2\) | 2. | \(2:1\) |
3. | \(4:1\) | 4. | \(1:1\) |
The Young's modulus of a wire is numerically equal to the stress at a point when:
1. | the strain produced in the wire is equal to unity. |
2. | the length of the wire gets doubled. |
3. | the length increases by \(100\%\). |
4. | All of these |
A metallic rope of diameter \(1~ \text{mm}\) breaks at \(10 ~\text{N}\) force. If the wire of the same material has a diameter of \(2~\text{mm}\), then the breaking force is:
1. | \(2.5~\text{N}\) | 2. | \(5~\text{N}\) |
3. | \(20~\text{N}\) | 4. | \(40~\text{N}\) |
In the CGS system, Young's modulus of a steel wire is \(2\times 10^{12}\) dyne/cm2. To double the length of a wire of unit cross-section area, the force required is:
1. \(4\times 10^{6}\) dynes
2. \(2\times 10^{12}\) dynes
3. \(2\times 10^{12}\) newtons
4. \(2\times 10^{8}\) dynes
On applying stress of \(20 \times 10^{8}~\text{N/m}^2\), the length of a perfectly elastic wire is doubled. It's Young’s modulus will be:
1. | \(40 \times 10^{8}~\text{N/m}^2\) | 2. | \(20 \times 10^{8}~\text{N/m}^2\) |
3. | \(10 \times 10^{8}~\text{N/m}^2\) | 4. | \(5 \times 10^{8}~\text{N/m}^2\) |
The area of cross-section of a wire of length \(1.1\) m is \(1\) mm2. It is loaded with mass of \(1\) kg. If Young's modulus of copper is \(1.1\times10^{11}\) N/m2, then the increase in length will be: (If \(g = 10~\text{m/s}^2)\)
1. | \(0.01\) mm | 2. | \(0.075\) mm |
3. | \(0.1\) mm | 4. | \(0.15\) mm |