
A man grows into a giant such that his linear dimensions increase by a factor of $9.$Assuming that his density remains same, the stress in the leg well change by a factor of:
A. $\dfrac{1}{{81}}$
B. $9$
C. $\dfrac{1}{9}$
D. $81$
Answer
585.3k+ views
Hint:Here,we are going to apply the concept of stress and it is defined as the force applied per unit area.If the area is high then stress become less and when is low then stress will be maximum provided that force applied on it remains constant.
Complete step by step answer:
We know that stress is defined as the ratio of force applied to the area of cross-section. Consider the height and width of the man as a linear dimension as \[l\] and \[b\].
Then, initial stress$ = s = \dfrac{F}{{l \times b}}$ . . . (1)
Now, the dimensions increase by the factor of $9$ i.e. ${l^1} = 9l$ and ${b^1} = 9b$
Hence, now stress ${s^1} = \dfrac{{{F^1}}}{{{l^1} \times {L^1}}}$
$ \Rightarrow {s^1} = \dfrac{{{F^1}}}{{9l \times 9b}}$
$ \Rightarrow {s^1} = \dfrac{{{F^1}}}{{81lb}}$ . . . (2)
Dividing equation (2) by equation (1) we get
$\dfrac{{{S^1}}}{S} = \dfrac{{\dfrac{{{F^1}}}{{81lb}}}}{{\dfrac{F}{{lb}}}}$
$ = \dfrac{{{F^1}}}{{81lb}} \times \dfrac{{lb}}{F}$
$ \Rightarrow \dfrac{{{S^1}}}{S} = \dfrac{{{F^1}}}{{81lb}}$ . . . (3)
Now, $F = ma = mg.$
Since, the force 1) due to the acceleration due to gravity we know that, volume density is mass per unit volume.
i.e. $e = \dfrac{m}{V}$
$ \Rightarrow m = eV$
$ \Rightarrow F = eVg$
Now, since linear dimensions have changed by a factor of $9$ volume would change by a factor of ${9^3}.$
$\therefore {F^1} = eVg \times {9^3}.$
Putting these volumes is equation (3), we get
$\dfrac{{{S^1}}}{S} = \dfrac{{eVg \times {9^1}}}{{81 \times eVg}}$
$ \Rightarrow \dfrac{{{S^1}}}{S} = 9(\therefore 81 = {9^2})$
$ \Rightarrow {S^1} = 95$
Therefore, the above explanation, the correct option is (B) 9.
Note:We could solve this question just by observing the options. Clearly, finding the size of mass is increasing, the stress is his leg must increase. Therefore, (A) and (C) cannot be the options.
Since, all the dimensions are increasing equally, by the factor of $9$ and density is the same. Then, stress should also increase by the factor of $9.$ Therefore, option (B) should be the answer.
Complete step by step answer:
We know that stress is defined as the ratio of force applied to the area of cross-section. Consider the height and width of the man as a linear dimension as \[l\] and \[b\].
Then, initial stress$ = s = \dfrac{F}{{l \times b}}$ . . . (1)
Now, the dimensions increase by the factor of $9$ i.e. ${l^1} = 9l$ and ${b^1} = 9b$
Hence, now stress ${s^1} = \dfrac{{{F^1}}}{{{l^1} \times {L^1}}}$
$ \Rightarrow {s^1} = \dfrac{{{F^1}}}{{9l \times 9b}}$
$ \Rightarrow {s^1} = \dfrac{{{F^1}}}{{81lb}}$ . . . (2)
Dividing equation (2) by equation (1) we get
$\dfrac{{{S^1}}}{S} = \dfrac{{\dfrac{{{F^1}}}{{81lb}}}}{{\dfrac{F}{{lb}}}}$
$ = \dfrac{{{F^1}}}{{81lb}} \times \dfrac{{lb}}{F}$
$ \Rightarrow \dfrac{{{S^1}}}{S} = \dfrac{{{F^1}}}{{81lb}}$ . . . (3)
Now, $F = ma = mg.$
Since, the force 1) due to the acceleration due to gravity we know that, volume density is mass per unit volume.
i.e. $e = \dfrac{m}{V}$
$ \Rightarrow m = eV$
$ \Rightarrow F = eVg$
Now, since linear dimensions have changed by a factor of $9$ volume would change by a factor of ${9^3}.$
$\therefore {F^1} = eVg \times {9^3}.$
Putting these volumes is equation (3), we get
$\dfrac{{{S^1}}}{S} = \dfrac{{eVg \times {9^1}}}{{81 \times eVg}}$
$ \Rightarrow \dfrac{{{S^1}}}{S} = 9(\therefore 81 = {9^2})$
$ \Rightarrow {S^1} = 95$
Therefore, the above explanation, the correct option is (B) 9.
Note:We could solve this question just by observing the options. Clearly, finding the size of mass is increasing, the stress is his leg must increase. Therefore, (A) and (C) cannot be the options.
Since, all the dimensions are increasing equally, by the factor of $9$ and density is the same. Then, stress should also increase by the factor of $9.$ Therefore, option (B) should be the answer.
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