
Two blocks of masses $ m $ and $ M>2m $ are connected by means of a metal wire of cross-sectional area A passing over a frictionless fixed pulley as shown in figure. The system is then released.
If $ m=1kg,A=8\times {{10}^{-9}}{{m}^{2}} $ , the breaking stress $ =2\times {{10}^{9}}N{{m}^{-2}} $ and $ g=10m{{s}^{-2}} $ , the maximum value of $ M $ for which the wire will not break is
a. $ 4kg $
b. $ 6kg $
c. $ 8kg $
d. $ 10kg $
Answer
545.4k+ views
Hint :Breaking stress is the maximum value of stress that a specimen can withstand without being failure.
Value of stress $ =\dfrac{Tension}{cross-\sec tonal\,area}=\dfrac{T}{A} $ .
Complete Step By Step Answer:
Let say after released blocks move with acceleration a and tension in wire $ =T $
from the $ F.B.D $ of mass $ m $ we can say
$ T-mg=ma...................\left( i \right) $
from the $ F.B.D $ of mass $ M $ we can say
$ Mg-T=Ma..................\left( ii \right) $
by adding the equation (i) and (ii)
$ Mg-mg=\left( m+M \right)a $
$ a=\left( \dfrac{M-m}{m+M} \right)g $
By putting thin equation (i)
$ T=mg+m\left( \dfrac{M-m}{m+M} \right)g $
Given that $ m=1kg $
$ T=1\times 10+\dfrac{1\left( M-1 \right)}{1+M}\left( 10 \right)=\left( \dfrac{20m}{M+1} \right)N..............\left( iii \right) $
Now we also know that
Stress $ =\dfrac{T}{A} $
And for the condition that wire will not break following coding must be followed stress in wire $ \le $ breaking stress
In question given that breaking stress $ =2\times {{10}^{9}}{\scriptstyle{}^{n}/{}_{{{m}^{2}}}} $
Therefore $ \dfrac{T}{A}\le 2\times {{10}^{9}}{}^{N}/{}_{m} $
By putting the value of $ T $ and $ A $
$ \dfrac{\left( \tfrac{20M}{1+M} \right)}{8\times {{10}^{-9}}}\le 2\times {{10}^{9}} $
$ \Rightarrow \dfrac{20M}{1+M}\le 16 $
$ \Rightarrow 20M\le 16+16M $
$ \Rightarrow 4M\le 16 $
$ \Rightarrow M\le 4kg $
So $ M $ should be less than equal to $ 4kg $ in order to ensure the wire will not break.
Note :
Stress is not tension on the unit area of the cross section, but it is the internal resistance produced against the external force exerted on an object. The value of the stress is the ratio of external force and area.
Value of stress $ =\dfrac{Tension}{cross-\sec tonal\,area}=\dfrac{T}{A} $ .
Complete Step By Step Answer:
Let say after released blocks move with acceleration a and tension in wire $ =T $
from the $ F.B.D $ of mass $ m $ we can say
$ T-mg=ma...................\left( i \right) $
from the $ F.B.D $ of mass $ M $ we can say
$ Mg-T=Ma..................\left( ii \right) $
by adding the equation (i) and (ii)
$ Mg-mg=\left( m+M \right)a $
$ a=\left( \dfrac{M-m}{m+M} \right)g $
By putting thin equation (i)
$ T=mg+m\left( \dfrac{M-m}{m+M} \right)g $
Given that $ m=1kg $
$ T=1\times 10+\dfrac{1\left( M-1 \right)}{1+M}\left( 10 \right)=\left( \dfrac{20m}{M+1} \right)N..............\left( iii \right) $
Now we also know that
Stress $ =\dfrac{T}{A} $
And for the condition that wire will not break following coding must be followed stress in wire $ \le $ breaking stress
In question given that breaking stress $ =2\times {{10}^{9}}{\scriptstyle{}^{n}/{}_{{{m}^{2}}}} $
Therefore $ \dfrac{T}{A}\le 2\times {{10}^{9}}{}^{N}/{}_{m} $
By putting the value of $ T $ and $ A $
$ \dfrac{\left( \tfrac{20M}{1+M} \right)}{8\times {{10}^{-9}}}\le 2\times {{10}^{9}} $
$ \Rightarrow \dfrac{20M}{1+M}\le 16 $
$ \Rightarrow 20M\le 16+16M $
$ \Rightarrow 4M\le 16 $
$ \Rightarrow M\le 4kg $
So $ M $ should be less than equal to $ 4kg $ in order to ensure the wire will not break.
Note :
Stress is not tension on the unit area of the cross section, but it is the internal resistance produced against the external force exerted on an object. The value of the stress is the ratio of external force and area.
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