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A wheel makes 20 revolutions per hour. T...

A wheel makes 20 revolutions per hour. The radians turns through 25 minutes is __________.

A

`(50pi^(c))/(7)`

B

`(250pi^(c))/(3)`

C

`(150pi^(c))/(7)`

D

`(50pi^(c))/(3)`

Text Solution

AI Generated Solution

The correct Answer is:
To find the radians turned through by a wheel that makes 20 revolutions per hour in 25 minutes, we can follow these steps: ### Step 1: Calculate the number of revolutions in 25 minutes. Given that the wheel makes 20 revolutions in 1 hour (which is 60 minutes), we can find the number of revolutions in 1 minute: \[ \text{Revolutions per minute} = \frac{20 \text{ revolutions}}{60 \text{ minutes}} = \frac{1}{3} \text{ revolutions per minute} \] Now, to find the number of revolutions in 25 minutes: \[ \text{Revolutions in 25 minutes} = \left(\frac{1}{3} \text{ revolutions per minute}\right) \times 25 \text{ minutes} = \frac{25}{3} \text{ revolutions} \] ### Step 2: Convert revolutions to radians. We know that 1 revolution is equal to \(2\pi\) radians. Therefore, to find the total radians for \(\frac{25}{3}\) revolutions: \[ \text{Radians} = \left(\frac{25}{3} \text{ revolutions}\right) \times (2\pi \text{ radians per revolution}) = \frac{25 \times 2\pi}{3} = \frac{50\pi}{3} \text{ radians} \] ### Final Answer: The radians turned through in 25 minutes is \(\frac{50\pi}{3}\). ---

To find the radians turned through by a wheel that makes 20 revolutions per hour in 25 minutes, we can follow these steps: ### Step 1: Calculate the number of revolutions in 25 minutes. Given that the wheel makes 20 revolutions in 1 hour (which is 60 minutes), we can find the number of revolutions in 1 minute: \[ \text{Revolutions per minute} = \frac{20 \text{ revolutions}}{60 \text{ minutes}} = \frac{1}{3} \text{ revolutions per minute} \] ...
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PEARSON IIT JEE FOUNDATION-TRIGONOMETRY-Level 1
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  2. Write the value of sintheta cos(90^(@)-theta)+costheta sin(90^(@)-the...

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  3. A wheel makes 20 revolutions per hour. The radians turns through 25 mi...

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  4. (sin^(4)theta-cos^(4)theta)/(sin^(2)-cos^(2)theta) = .

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  5. Simplified expression of (sectheta+tantheta)(1-sintheta) is .

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  6. If a=sectheta-tantheta and b= sectheta+tantheta, then

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  7. If secalpha+tanalpha=m, " then " sec^(4)alpha-tan^(4)alpha-2sec alpha...

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  8. If sin^(4)A-cos^(4)A=1,then (A//2)is .(0ltAle90^(@)).

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  9. The value of tan 15^(@) tan20^(@) tan 70^(@) tan75 is

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  10. In a DeltaABC, tan((A+C)/(2)) = .

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  11. If tan (A-30^(@))=2-sqrt3, then find A.

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  12. If sin^(4)theta-cos^(4)=k^(4), then sin^(2)theta-cos^(2)theta is .

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  13. (tan^(3)theta-1)/(tantheta-1)= .

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  14. For all values of theta,1+costheta can be.

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  15. If sin3theta=cos(theta-6^(@))," where "3theta and (theta-6^(@)) are ac...

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  16. (cosecA-sinA)(secA-cosA)(tanA+cotA) = .

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  17. Ifx=a(cosectheta+cottheta)and y=b(cottheta-cosectheta), then

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  18. The value of (cos^(4)x+cos^(2)xsin^(2)x+sin^(2)x)/(cos^(2)x+sin^(2)xc...

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  19. (1)/(1+sintheta)+(1)/(1-sintheta) is equal to .

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  20. if tan(alpha+beta)=(1)/(2) and tanalpha=(1)/(3), then tanbeta=.

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