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Wheels of diameters 7 cm and 14 cm start rolling simultaneously from X and Y, which are 1980 cm apart, towards each other in opposite directions. Both of them make the same number of revolutions per second. If both of them meet after 10 seconds, the speed of the smaller wheel is :
Answer & Solution
Correct Answer:
Option
A
Let each wheel x revolutions per second
Then,
$$\left[ {\left( {2\pi \times \frac{7}{2} \times x} \right) + \left( {2\pi \times 7 \times x} \right)} \right]$$ $$ \times 10$$ $$ = 1980$$
$$ \Rightarrow \left( {\frac{{22}}{7} \times 7 \times x} \right) + $$ $$\left( {2 \times \frac{{22}}{7} \times 7 \times x} \right)$$ $$ = 198$$
$$\eqalign{ & \Rightarrow 66x = 198 \cr & \Rightarrow x = 3 \cr} $$
Distance moved by smaller wheel in 3 revolutions :
$$\eqalign{ & = \left( {2 \times \frac{{22}}{7} \times \frac{7}{2} \times 3} \right)cm \cr & = 66\,cm \cr} $$
∴ Speed of smaller wheel :
$$\eqalign{ & = \frac{{66}}{3}cm/\sec \cr & = 22\,cm/\sec \cr} $$
Then,
$$\left[ {\left( {2\pi \times \frac{7}{2} \times x} \right) + \left( {2\pi \times 7 \times x} \right)} \right]$$ $$ \times 10$$ $$ = 1980$$
$$ \Rightarrow \left( {\frac{{22}}{7} \times 7 \times x} \right) + $$ $$\left( {2 \times \frac{{22}}{7} \times 7 \times x} \right)$$ $$ = 198$$
$$\eqalign{ & \Rightarrow 66x = 198 \cr & \Rightarrow x = 3 \cr} $$
Distance moved by smaller wheel in 3 revolutions :
$$\eqalign{ & = \left( {2 \times \frac{{22}}{7} \times \frac{7}{2} \times 3} \right)cm \cr & = 66\,cm \cr} $$
∴ Speed of smaller wheel :
$$\eqalign{ & = \frac{{66}}{3}cm/\sec \cr & = 22\,cm/\sec \cr} $$
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