Uniform rod of mass m, length l, area of cross-section A has Young’s modulus Y.If it is hanged vertically, elongation under its own weight will be
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Solution
\(Y==\frac{F\iota }{A\Delta \iota }\Rightarrow \Delta \iota =\frac{F\iota }{YA}=\frac{mg\iota }{YA}\)
Which one of the following affects the elasticity of a substance ?
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Solution
The elasticity of a material depends upon the temperature of the material. Hammering & annealing reduces elastic property of a substance.
According to Hooke’s law of elasticity, if stress is increased, then the ratio of stress to strain
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Solution
The ratio of stress to strain is always constant. If stress is increased, strain will also increase so that their ratio remains constant.
The length of an iron wire is L and area of corss-section is A. The increase in length is l on applying the force F on its two ends. Which of the statement is correct?
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Solution
\(\iota =\frac{FL}{YA}\Rightarrow \iota \alpha \frac{1}{A}\)
A and B are two wires. The radius of A is twice that of B.They are stretched by the same load. Then the stress on B is
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Solution
\(Stress =\frac{Force}{Area}∴Stress\alpha \frac{1}{\pi r^{2}}\)
\(\frac{S_{B}}{S_{A}}=\left ( \frac{r_{A}}{r_{B}} \right )^{2}=(2)^{2}\Rightarrow S_{B}=4S_{A}\)
A force of 103 newton, stretches the length of a hanging wire by 1 millimetre. The force required to stretch awire of same material and length but having four times the diameter by 1 millimetre is
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Solution
\(F=Y\times A\times \frac{\iota }{L}\Rightarrow F\alpha r^{2}\) (Y, l and and L are constant)If diameter is made four times then force required will be 16 times, i.e., 16 × 103 N
In case of steel wire (or a metal wire), the limit is reached when
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Solution
According to Hooke's Law, within the elastic limits stress is directly proportional to strain.
Fora given material, the Young’s modulus is 2. 4 times that of rigidity modulus. Its Poisson’s ratio is
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Solution
\(Y=2\eta (1+\sigma )\)
\(2.4\eta =2\eta (1+\sigma )\Rightarrow 1.2=1+\sigma \Rightarrow \sigma =0.2\)
Which of the following relation is true ?
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Solution
\(Y=2\eta (1+\sigma )\Rightarrow \sigma =\frac{0.5Y-\eta }{\eta }\)
For a constant hydraulic stress on an object, the fractional change in the object volume \(\left ( \frac{\Delta V}{V} \right )\) and its bulk modulus(B)are related as
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Solution
\(B=\frac{\Delta p}{\Delta v/v}\Rightarrow \frac{1}{B}\alpha \frac{\Delta v}{v}\) [Δp = constant]