A force F is needed to break a copper wire having radius R. The force needed to break a copper wire of radius 2R will be.
A. $\dfrac{F}{2}$
B. 2F
C. 4F
D. $\dfrac{F}{4}$
Answer
585.6k+ views
Hint: Force: It is a push or pull on an object. It is denoted by ‘F’. SI unit is ‘N’ or Newton. There are four fundamental forces or we say basic forces they are gravitational force, electromagnetic force, weak nuclear force, strong nuclear force. Gravity is the weakest force and strong nuclear force is the strongest force. As we know that breaking force is directly proportional to the cross section of the object.
$ \Rightarrow F \propto \pi {r^2}$
Complete step by step solution:
According to question
Given,
r=radius of wire
A=cross section of wire
F=breaking force
Breaking force Cross section area
$ \Rightarrow F \propto \pi {r^2}$
‘r’ is the radius of given object
For radius ‘R’ breaking force is ‘F’
\[ \Rightarrow F = \pi {R^2}\] …(1)
For radius ’2R’ breaking force will be
\[ \Rightarrow {F_2} = \pi {(2R)^2}\]
\[ \Rightarrow {F_2} = \pi 4{R^2}\] …(2)
From equation (1) and (2)
$ \Rightarrow \dfrac{{{F_2}}}{F} = \dfrac{{\pi 4{R^2}}}{{\pi {R^2}}}$
$ \Rightarrow {F_2} = 4F$
So the answer is (C) 4F.
Note: That produces acceleration in the body or which it acts (a) it can change the speed of a body. (b) It can change the direction of the body. (c) It can change the shape of the body. Force is an external agent. Force is vector quantity; it has direction as well as magnitude.
$ \Rightarrow F \propto \pi {r^2}$
Complete step by step solution:
According to question
Given,
r=radius of wire
A=cross section of wire
F=breaking force
Breaking force Cross section area
$ \Rightarrow F \propto \pi {r^2}$
‘r’ is the radius of given object
For radius ‘R’ breaking force is ‘F’
\[ \Rightarrow F = \pi {R^2}\] …(1)
For radius ’2R’ breaking force will be
\[ \Rightarrow {F_2} = \pi {(2R)^2}\]
\[ \Rightarrow {F_2} = \pi 4{R^2}\] …(2)
From equation (1) and (2)
$ \Rightarrow \dfrac{{{F_2}}}{F} = \dfrac{{\pi 4{R^2}}}{{\pi {R^2}}}$
$ \Rightarrow {F_2} = 4F$
So the answer is (C) 4F.
Note: That produces acceleration in the body or which it acts (a) it can change the speed of a body. (b) It can change the direction of the body. (c) It can change the shape of the body. Force is an external agent. Force is vector quantity; it has direction as well as magnitude.
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