
A wire can sustain a weight of $ 15kg $ . If it is cut into four equal parts, then each part can sustain a weight
(A) $ 5kg $
(B) $ 45kg $
(C) $ 15kg $
(D) $ 30kg $
Answer
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Hint : To solve this question, we need to evaluate the value of the breaking stress of the wire by using the basic formula of stress. Then equating this value for the two cases given in the question, we can get the final answer.
Formula used: The formula used to solve this question is given by
$ \sigma = \dfrac{F}{A} $ , here $ \sigma $ is the stress due to a force of $ F $ on a surface of area $ A $ .
Complete step by step answer
When a weight is suspended on a hanging wire, a longitudinal force acts on it. Due to this, the longitudinal stress is developed in the wire. There is a limit to which a material can sustain a longitudinal stress, beyond which the wire breaks. This breaking stress is the property of the material of the wire. So we have to determine the value of this breaking stress of the material of the wire given.
Let the original length of the wire be $ L $ and its cross sectional area be $ A $ .
We know that the stress is given by
$ \sigma = \dfrac{F}{A} $
The maximum force sustained by the wire is equal to the weight of $ 15kg $ . So the breaking stress is given by
$ \sigma = \dfrac{{15g}}{A} $ ..........................(1)
Now, the wire is cut into four equal parts, so that the length of each part becomes $ \dfrac{L}{4} $ . But the cross sectional area of each of the four parts will remain the same, that is, $ A $ .
Let the weight sustained by each part be $ m $ . So the breaking force for the part is given by
$ {F_2} = mg $
And the corresponding breaking stress is given by
$ \sigma = \dfrac{{mg}}{A} $ ..........................(2)
Since the breaking stress is constant for the material of the wire, so we can equate (1) and (2) to get
$ \dfrac{{mg}}{A} = \dfrac{{15g}}{A} $
$ \Rightarrow m = 15kg $
Thus, each of the four parts of the wire can sustain the same weight of $ 15kg $ as sustained by the original wire.
Hence, the correct answer is option C.
Note
While solving this question, we have assumed that the given wire is cut longitudinally. The direction of cutting is not mentioned in the question; also the thickness of a wire is negligible compared to its length. So we have made this assumption.
Formula used: The formula used to solve this question is given by
$ \sigma = \dfrac{F}{A} $ , here $ \sigma $ is the stress due to a force of $ F $ on a surface of area $ A $ .
Complete step by step answer
When a weight is suspended on a hanging wire, a longitudinal force acts on it. Due to this, the longitudinal stress is developed in the wire. There is a limit to which a material can sustain a longitudinal stress, beyond which the wire breaks. This breaking stress is the property of the material of the wire. So we have to determine the value of this breaking stress of the material of the wire given.
Let the original length of the wire be $ L $ and its cross sectional area be $ A $ .
We know that the stress is given by
$ \sigma = \dfrac{F}{A} $
The maximum force sustained by the wire is equal to the weight of $ 15kg $ . So the breaking stress is given by
$ \sigma = \dfrac{{15g}}{A} $ ..........................(1)
Now, the wire is cut into four equal parts, so that the length of each part becomes $ \dfrac{L}{4} $ . But the cross sectional area of each of the four parts will remain the same, that is, $ A $ .
Let the weight sustained by each part be $ m $ . So the breaking force for the part is given by
$ {F_2} = mg $
And the corresponding breaking stress is given by
$ \sigma = \dfrac{{mg}}{A} $ ..........................(2)
Since the breaking stress is constant for the material of the wire, so we can equate (1) and (2) to get
$ \dfrac{{mg}}{A} = \dfrac{{15g}}{A} $
$ \Rightarrow m = 15kg $
Thus, each of the four parts of the wire can sustain the same weight of $ 15kg $ as sustained by the original wire.
Hence, the correct answer is option C.
Note
While solving this question, we have assumed that the given wire is cut longitudinally. The direction of cutting is not mentioned in the question; also the thickness of a wire is negligible compared to its length. So we have made this assumption.
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