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NEET-XII-Physics

14: Some Mechanical Properties of Matter

with Solutions - page 5
Qstn# iv-6-a Prvs-QstnNext-Qstn
  • #6-a
    Find the maximum load that can be put on the hanger without breaking a wire. Which wire will break first if the load is increased? (b) Repeat the above part if m1 = 10 kg and m2 = 36 kg. (b) Repeat the above part if m1 = 10 kg and m2 = 36 kg.
    Ans : Given:
    `` \,\mathrm{\,Breaking\,}\,\mathrm{\,stress\,}\,\mathrm{\,of\,}\,\mathrm{\,wire\,}=8\times {10}^{8}\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}``
    `` \,\mathrm{\,Area\,}\,\mathrm{\,of\,}\,\mathrm{\,cross\,}-\,\mathrm{\,section\,}\,\mathrm{\,of\,}\,\mathrm{\,upper\,}\,\mathrm{\,wire\,}\left({A}_{\,\mathrm{\,u\,}}\right)=0.006{\,\mathrm{\,cm\,}}^{2}=6\times {10}^{-7}\,\mathrm{\,m\,}``
    `` \,\mathrm{\,Area\,}\,\mathrm{\,of\,}\,\mathrm{\,cross\,}-\,\mathrm{\,section\,}\,\mathrm{\,of\,}\,\mathrm{\,lower\,}\,\mathrm{\,wire\,}\left({A}_{\,\mathrm{\,l\,}}\right)=0.003{\,\mathrm{\,cm\,}}^{2}=3\times {10}^{-7}\,\mathrm{\,m\,}``
    `` {m}_{1}=10\,\mathrm{\,kg\,},{m}_{2}=20\,\mathrm{\,kg\,}``
    Tension in lower wire `` {T}_{l}={m}_{1}g+w``
    Here: g is the acceleration due to gravity
    w is the load
    ∴ Stress in lower wire`` =\frac{{\,\mathrm{\,T\,}}_{l}}{{\,\mathrm{\,A\,}}_{l}}=\frac{{m}_{1}g+w}{{\,\mathrm{\,A\,}}_{l}}``
    `` \Rightarrow \frac{{m}_{1}g+w}{{A}_{l}}=8\times {10}^{8}``
    `` \Rightarrow w=\left[\left(8\times {10}^{8}\right)\times \left(3\times {10}^{-7}\right)\right]-100``
    `` ``
    `` \Rightarrow w=140\,\mathrm{\,N\,}\,\mathrm{\,or\,}14\,\mathrm{\,kg\,}``
    Now, tension in upper wire `` {T}_{2}={m}_{1}g+{m}_{2}g+w``
    ∴ Stress in upper wire`` =\frac{{\,\mathrm{\,T\,}}_{u}}{{\,\mathrm{\,A\,}}_{u}}=\frac{{m}_{\mathit{2}}g+{m}_{1}g+w}{{\,\mathrm{\,A\,}}_{u}}``
    `` \Rightarrow \frac{{m}_{2}g+{m}_{1}g+w}{{\,\mathrm{\,A\,}}_{u}}=8\times {10}^{8}``
    `` \Rightarrow w=180\,\mathrm{\,N\,}\,\mathrm{\,or\,}18\,\mathrm{\,kg\,}``
    For the same breaking stress, the maximum load that can be put is 140 N or 14 kg. The lower wire will break first if the load is increased. (b) `` \,\mathrm{\,If\,}{m}_{1}=10\,\mathrm{\,kg\,}\,\mathrm{\,and\,}{m}_{2}=36\,\mathrm{\,kg\,}``:
    Tension in lower wire `` {T}_{l}={m}_{1}g+w``
    Here: g is the acceleration due to gravity
    w is the load
    ∴ Stress in lower wire:
    `` \Rightarrow \frac{{\,\mathrm{\,T\,}}_{l}}{{\,\mathrm{\,A\,}}_{l}}=\frac{{m}_{1}g+w}{{\,\mathrm{\,A\,}}_{l}}=8\times {10}^{5}``
    `` \Rightarrow w=140\,\mathrm{\,N\,}``
    Now, tension in upper wire `` {T}_{2}={m}_{1}g+{m}_{2}g+w``
    ∴ Stress in upper wire:
    `` \Rightarrow \frac{{\,\mathrm{\,T\,}}_{u}}{{\,\mathrm{\,A\,}}_{u}}=\frac{{m}_{2}g+{m}_{1}g+w}{{\,\mathrm{\,A\,}}_{u}}=8\times {10}^{5}``
    `` \Rightarrow w=20\,\mathrm{\,N\,}``
    For the same breaking stress, the maximum load that can be put is 20 N or 2 kg. The upper wire will break first if the load is increased.
    Page No 300: (b) `` \,\mathrm{\,If\,}{m}_{1}=10\,\mathrm{\,kg\,}\,\mathrm{\,and\,}{m}_{2}=36\,\mathrm{\,kg\,}``:
    Tension in lower wire `` {T}_{l}={m}_{1}g+w``
    Here: g is the acceleration due to gravity
    w is the load
    ∴ Stress in lower wire:
    `` \Rightarrow \frac{{\,\mathrm{\,T\,}}_{l}}{{\,\mathrm{\,A\,}}_{l}}=\frac{{m}_{1}g+w}{{\,\mathrm{\,A\,}}_{l}}=8\times {10}^{5}``
    `` \Rightarrow w=140\,\mathrm{\,N\,}``
    Now, tension in upper wire `` {T}_{2}={m}_{1}g+{m}_{2}g+w``
    ∴ Stress in upper wire:
    `` \Rightarrow \frac{{\,\mathrm{\,T\,}}_{u}}{{\,\mathrm{\,A\,}}_{u}}=\frac{{m}_{2}g+{m}_{1}g+w}{{\,\mathrm{\,A\,}}_{u}}=8\times {10}^{5}``
    `` \Rightarrow w=20\,\mathrm{\,N\,}``
    For the same breaking stress, the maximum load that can be put is 20 N or 2 kg. The upper wire will break first if the load is increased.
    Page No 300:
  • #6-b
    Repeat the above part if m1 = 10 kg and m2 = 36 kg.
    Ans : `` \,\mathrm{\,If\,}{m}_{1}=10\,\mathrm{\,kg\,}\,\mathrm{\,and\,}{m}_{2}=36\,\mathrm{\,kg\,}``:
    Tension in lower wire `` {T}_{l}={m}_{1}g+w``
    Here: g is the acceleration due to gravity
    w is the load
    ∴ Stress in lower wire:
    `` \Rightarrow \frac{{\,\mathrm{\,T\,}}_{l}}{{\,\mathrm{\,A\,}}_{l}}=\frac{{m}_{1}g+w}{{\,\mathrm{\,A\,}}_{l}}=8\times {10}^{5}``
    `` \Rightarrow w=140\,\mathrm{\,N\,}``
    Now, tension in upper wire `` {T}_{2}={m}_{1}g+{m}_{2}g+w``
    ∴ Stress in upper wire:
    `` \Rightarrow \frac{{\,\mathrm{\,T\,}}_{u}}{{\,\mathrm{\,A\,}}_{u}}=\frac{{m}_{2}g+{m}_{1}g+w}{{\,\mathrm{\,A\,}}_{u}}=8\times {10}^{5}``
    `` \Rightarrow w=20\,\mathrm{\,N\,}``
    For the same breaking stress, the maximum load that can be put is 20 N or 2 kg. The upper wire will break first if the load is increased.
    Page No 300: