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The radius of a wire is decreased to one-third. If volume remains the same, length will increase by :1). 1.5 times2). 3 times3). 6 times4). 9 times |
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Answer» Let the initial radius and length of the wire be R and l respectively. Now, radius changes to r/3, let the new length be $(\;{l_1})$. ? Volume does not change, $(\PI {r^2}l = \pi {\left( {\frac{r}{3}} \RIGHT)^2}{l_1})$ ⇒ $({r^2}l = \frac{{{r^2}}}{9} \times {l_1})$ ⇒ l1 = 9l ∴ Length is increased by 9 times. |
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