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Standard XII

Business Studies

Manager & Management

A man does a ...

Question

A man does a given amount of work in $10 s e c$ . Another man does the same amount of work in $20 s e c$ . The ratio of the output power of first man to the second man is $\frac{x}{y}$ Find $(x - y)$ .

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Solution

Formula Used: $P = \frac{w}{t}$

Power $= \frac{w}{t}$
If $W$ is constant, then $P \propto \frac{1}{t}$

i.e. $\frac{P_{1}}{P_{2}} = \frac{t_{2}}{t_{1}} = \frac{20}{10} = \frac{2}{1}$

As per question $x - y = 1$

Final Answer: $1$

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Manager & Management

Standard XII Business Studies

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A man does a given amount of work in 10 sec. Another man does the same amount of work in 20 sec. The ratio of the output power of first man to the second man is xy Find (x−y).

Formula Used: P=wt Power =wt If W is constant, then P∝1t i.e. P1P2=t2t1=2010=21 As per question x−y=1 Final Answer: 1

A man does a given amount of work in $10 s e c$ . Another man does the same amount of work in $20 s e c$ . The ratio of the output power of first man to the second man is $\frac{x}{y}$ Find $(x - y)$ .

Formula Used: $P = \frac{w}{t}$

Power $= \frac{w}{t}$
If $W$ is constant, then $P \propto \frac{1}{t}$

i.e. $\frac{P_{1}}{P_{2}} = \frac{t_{2}}{t_{1}} = \frac{20}{10} = \frac{2}{1}$

As per question $x - y = 1$

Final Answer: $1$