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NEET Physics Wave Optics / तरंग प्रकाशिकी NEET PYQ-Wave Optics

  • question_answer
    The Young's modulus of steel is twice that of brass. Two wires of same length and of same area of cross-section, one of steel and another of brass are suspended from the same roof. It we want the lower ends of the wires to be at the same level, then the weight added to the steel and brass wires must be in the ratio of                                          [NEET (Re) 2015]

    A)  1 : 2   

    B)       2 : 1   

    C)  4 : 1        

    D)              1 : 1

    Correct Answer: B

    Solution :

    Given,   \[{{Y}_{steel}}=2{{Y}_{brass}}\]
    and       \[{{L}_{s}}={{L}_{b}}\]
    and       \[{{A}_{a}}={{A}_{b}}\]
    such that \[\Delta {{L}_{s}}=\Delta {{L}_{b}}\]
    As we know, Young's modulus,
    \[Y=\frac{\text{stress}}{\text{strain}}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}=\frac{W\Delta L}{A\Delta L}\]
    So,       \[W=\frac{FA\Delta L}{L}\propto Y\]
    i.e.            \[\frac{{{W}_{s}}}{{{W}_{b}}}=\frac{{{Y}_{s}}}{{{Y}_{b}}}=\frac{2{{Y}_{b}}}{{{Y}_{b}}}=\frac{2}{1}\]
    \[\Rightarrow \]   \[2:1\]
    Thus, weight added to the steel and brass wires must be in the ratio of 2 : 1.


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