00. Elasticity, Stress, Strain and Hooke's law
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
Mechanical Properties of Solids

140936 Two wires of same material and same diameter have lengths in the ratio $2: 5$. They are stretched by same force. The ratio of work done in stretching them is:

1 $5: 2$
2 $2: 5$
3 $1: 3$
4 $3: 1$
Mechanical Properties of Solids

140937 Match the following
| | Column I | | Column II |
| :--- | :--- | :--- | :--- |
| (A) | Hooke's law | 1. | Tangential strain |
| (B) | Shearing strain | 2. | $\begin{array}{l}\text { Temporary loss of } \\ \text { elastic property }\end{array}$ |
| (C) | Bulk strain | 3. | Elastic limit |
| (D) | Elastic fatigue | 4. | Times |

1 2 (A) 1 (B) 4 (C) 3 (D)
2 3 (A) 4 (B) 1 (C) 2 (D)
3 3 (A) 1 (B) 4 (C) 2 (D)
4 1 (A) 2 (B) 3 (C) 4 (D)
Mechanical Properties of Solids

140938 Two wires of the same material and length but diameters in the ratio $1: 2$ are stretched by the same force. The elastic potential energy per unit volume for the wires, when stretched by the same force will be in the ratio.

1 $16: 1$
2 $1: 1$
3 $2: 1$
4 $4: 1$
Mechanical Properties of Solids

140939 When a force $F_{1}$ is applied on a metallic wire, the length of the wire is $L_{1}$. If a force $F_{2}$ is applied on the same wire, the length of the wire is $L_{2}$. The original length of the wire $L$ is

1 $\frac{\mathrm{L}_{1} \mathrm{~F}_{1}+\mathrm{L}_{2} \mathrm{~F}_{2}}{\mathrm{~F}_{1}+\mathrm{F}_{2}}$
2 $\frac{\mathrm{L}_{2}-\mathrm{L}_{1}}{\mathrm{~F}_{1}+\mathrm{F}_{2}}$
3 $\frac{\mathrm{F}_{1} \mathrm{~L}_{2}-\mathrm{F}_{2} \mathrm{~L}_{1}}{\mathrm{~F}_{1}-\mathrm{F}_{2}}$
4 $\frac{\mathrm{F}_{1} \mathrm{~L}_{1}-\mathrm{F}_{2} \mathrm{~L}_{2}}{\mathrm{~F}_{1}-\mathrm{F}_{2}}$
Mechanical Properties of Solids

140936 Two wires of same material and same diameter have lengths in the ratio $2: 5$. They are stretched by same force. The ratio of work done in stretching them is:

1 $5: 2$
2 $2: 5$
3 $1: 3$
4 $3: 1$
Mechanical Properties of Solids

140937 Match the following
| | Column I | | Column II |
| :--- | :--- | :--- | :--- |
| (A) | Hooke's law | 1. | Tangential strain |
| (B) | Shearing strain | 2. | $\begin{array}{l}\text { Temporary loss of } \\ \text { elastic property }\end{array}$ |
| (C) | Bulk strain | 3. | Elastic limit |
| (D) | Elastic fatigue | 4. | Times |

1 2 (A) 1 (B) 4 (C) 3 (D)
2 3 (A) 4 (B) 1 (C) 2 (D)
3 3 (A) 1 (B) 4 (C) 2 (D)
4 1 (A) 2 (B) 3 (C) 4 (D)
Mechanical Properties of Solids

140938 Two wires of the same material and length but diameters in the ratio $1: 2$ are stretched by the same force. The elastic potential energy per unit volume for the wires, when stretched by the same force will be in the ratio.

1 $16: 1$
2 $1: 1$
3 $2: 1$
4 $4: 1$
Mechanical Properties of Solids

140939 When a force $F_{1}$ is applied on a metallic wire, the length of the wire is $L_{1}$. If a force $F_{2}$ is applied on the same wire, the length of the wire is $L_{2}$. The original length of the wire $L$ is

1 $\frac{\mathrm{L}_{1} \mathrm{~F}_{1}+\mathrm{L}_{2} \mathrm{~F}_{2}}{\mathrm{~F}_{1}+\mathrm{F}_{2}}$
2 $\frac{\mathrm{L}_{2}-\mathrm{L}_{1}}{\mathrm{~F}_{1}+\mathrm{F}_{2}}$
3 $\frac{\mathrm{F}_{1} \mathrm{~L}_{2}-\mathrm{F}_{2} \mathrm{~L}_{1}}{\mathrm{~F}_{1}-\mathrm{F}_{2}}$
4 $\frac{\mathrm{F}_{1} \mathrm{~L}_{1}-\mathrm{F}_{2} \mathrm{~L}_{2}}{\mathrm{~F}_{1}-\mathrm{F}_{2}}$
Mechanical Properties of Solids

140936 Two wires of same material and same diameter have lengths in the ratio $2: 5$. They are stretched by same force. The ratio of work done in stretching them is:

1 $5: 2$
2 $2: 5$
3 $1: 3$
4 $3: 1$
Mechanical Properties of Solids

140937 Match the following
| | Column I | | Column II |
| :--- | :--- | :--- | :--- |
| (A) | Hooke's law | 1. | Tangential strain |
| (B) | Shearing strain | 2. | $\begin{array}{l}\text { Temporary loss of } \\ \text { elastic property }\end{array}$ |
| (C) | Bulk strain | 3. | Elastic limit |
| (D) | Elastic fatigue | 4. | Times |

1 2 (A) 1 (B) 4 (C) 3 (D)
2 3 (A) 4 (B) 1 (C) 2 (D)
3 3 (A) 1 (B) 4 (C) 2 (D)
4 1 (A) 2 (B) 3 (C) 4 (D)
Mechanical Properties of Solids

140938 Two wires of the same material and length but diameters in the ratio $1: 2$ are stretched by the same force. The elastic potential energy per unit volume for the wires, when stretched by the same force will be in the ratio.

1 $16: 1$
2 $1: 1$
3 $2: 1$
4 $4: 1$
Mechanical Properties of Solids

140939 When a force $F_{1}$ is applied on a metallic wire, the length of the wire is $L_{1}$. If a force $F_{2}$ is applied on the same wire, the length of the wire is $L_{2}$. The original length of the wire $L$ is

1 $\frac{\mathrm{L}_{1} \mathrm{~F}_{1}+\mathrm{L}_{2} \mathrm{~F}_{2}}{\mathrm{~F}_{1}+\mathrm{F}_{2}}$
2 $\frac{\mathrm{L}_{2}-\mathrm{L}_{1}}{\mathrm{~F}_{1}+\mathrm{F}_{2}}$
3 $\frac{\mathrm{F}_{1} \mathrm{~L}_{2}-\mathrm{F}_{2} \mathrm{~L}_{1}}{\mathrm{~F}_{1}-\mathrm{F}_{2}}$
4 $\frac{\mathrm{F}_{1} \mathrm{~L}_{1}-\mathrm{F}_{2} \mathrm{~L}_{2}}{\mathrm{~F}_{1}-\mathrm{F}_{2}}$
Mechanical Properties of Solids

140936 Two wires of same material and same diameter have lengths in the ratio $2: 5$. They are stretched by same force. The ratio of work done in stretching them is:

1 $5: 2$
2 $2: 5$
3 $1: 3$
4 $3: 1$
Mechanical Properties of Solids

140937 Match the following
| | Column I | | Column II |
| :--- | :--- | :--- | :--- |
| (A) | Hooke's law | 1. | Tangential strain |
| (B) | Shearing strain | 2. | $\begin{array}{l}\text { Temporary loss of } \\ \text { elastic property }\end{array}$ |
| (C) | Bulk strain | 3. | Elastic limit |
| (D) | Elastic fatigue | 4. | Times |

1 2 (A) 1 (B) 4 (C) 3 (D)
2 3 (A) 4 (B) 1 (C) 2 (D)
3 3 (A) 1 (B) 4 (C) 2 (D)
4 1 (A) 2 (B) 3 (C) 4 (D)
Mechanical Properties of Solids

140938 Two wires of the same material and length but diameters in the ratio $1: 2$ are stretched by the same force. The elastic potential energy per unit volume for the wires, when stretched by the same force will be in the ratio.

1 $16: 1$
2 $1: 1$
3 $2: 1$
4 $4: 1$
Mechanical Properties of Solids

140939 When a force $F_{1}$ is applied on a metallic wire, the length of the wire is $L_{1}$. If a force $F_{2}$ is applied on the same wire, the length of the wire is $L_{2}$. The original length of the wire $L$ is

1 $\frac{\mathrm{L}_{1} \mathrm{~F}_{1}+\mathrm{L}_{2} \mathrm{~F}_{2}}{\mathrm{~F}_{1}+\mathrm{F}_{2}}$
2 $\frac{\mathrm{L}_{2}-\mathrm{L}_{1}}{\mathrm{~F}_{1}+\mathrm{F}_{2}}$
3 $\frac{\mathrm{F}_{1} \mathrm{~L}_{2}-\mathrm{F}_{2} \mathrm{~L}_{1}}{\mathrm{~F}_{1}-\mathrm{F}_{2}}$
4 $\frac{\mathrm{F}_{1} \mathrm{~L}_{1}-\mathrm{F}_{2} \mathrm{~L}_{2}}{\mathrm{~F}_{1}-\mathrm{F}_{2}}$