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Question: Two blocks of masses \( m \) and \( M>2m \) are connected by means of a metal wire of cross-sectiona...

Two blocks of masses mm and M>2mM>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=8×10−9m2m=1kg,A=8\times {{10}^{-9}}{{m}^{2}} , the breaking stress =2×109Nm−2=2\times {{10}^{9}}N{{m}^{-2}} and g=10ms−2g=10m{{s}^{-2}} , the maximum value of MM for which the wire will not break is

a. 4kg4kg
b. 6kg6kg
c. 8kg8kg
d. 10kg10kg

Explanation

Solution

Hint : Breaking stress is the maximum value of stress that a specimen can withstand without being failure.
Value of stress =Tensioncross−sec⁡tonal area=TA=\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=T

from the F.B.DF.B.D of mass mm we can say
T−mg=ma...................(i)T-mg=ma...................\left( i \right)
from the F.B.DF.B.D of mass MM we can say
Mg−T=Ma..................(ii)Mg-T=Ma..................\left( ii \right)
by adding the equation (i) and (ii)
Mg−mg=(m+M)aMg-mg=\left( m+M \right)a
a=(M−mm+M)ga=\left( \dfrac{M-m}{m+M} \right)g
By putting thin equation (i)
T=mg+m(M−mm+M)gT=mg+m\left( \dfrac{M-m}{m+M} \right)g
Given that m=1kgm=1kg
T=1×10+1(M−1)1+M(10)=(20mM+1)N..............(iii)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 =TA=\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 =2×109n/m2=2\times {{10}^{9}}{\scriptstyle{}^{n}/{}_{{{m}^{2}}}}
Therefore TA≤2×109N/m\dfrac{T}{A}\le 2\times {{10}^{9}}{}^{N}/{}_{m}
By putting the value of TT and AA
(20M1+M)8×10−9≤2×109\dfrac{\left( \tfrac{20M}{1+M} \right)}{8\times {{10}^{-9}}}\le 2\times {{10}^{9}}
⇒20M1+M≤16\Rightarrow \dfrac{20M}{1+M}\le 16
⇒20M≤16+16M\Rightarrow 20M\le 16+16M
⇒4M≤16\Rightarrow 4M\le 16
⇒M≤4kg\Rightarrow M\le 4kg
So MM should be less than equal to 4kg4kg 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.