If the radius of a circle is increasing at the rate of 0.5 cm/s, then the rate of increase of its circumference is:
- (A) \( rac{2\pi}{3} \) cm/s
- (B) \( \pi \) cm/s
- (C) \( rac{4\pi}{3} \) cm/s
- (D) \( 2\pi \) cm/s
The Correct Option is B
Solution and Explanation
Step 1: Understand the given data
The radius of the circle is increasing at the rate of 0.5 cm/s. We need to find the rate of increase of the circumference.
Step 2: Recall the formula for circumference of a circle
The circumference C of a circle with radius r is given by:
C = 2πr
Step 3: Differentiate circumference with respect to time
Differentiate both sides with respect to time t:
dC/dt = 2π × dr/dt
Step 4: Substitute the given rate of change of radius
Given dr/dt = 0.5 cm/s, substitute into the differentiated equation:
dC/dt = 2π × 0.5 = π cm/s
Step 5: Conclusion
The rate of increase of the circumference is π cm/s.
Final Answer: (B) π cm/s
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In the following figure chord MN and chord RS intersect at point D. If RD = 15, DS = 4, MD = 8, find DN by completing the following activity:
Activity :
\(\therefore\) MD \(\times\) DN = \(\boxed{\phantom{SD}}\) \(\times\) DS \(\dots\) (Theorem of internal division of chords)
\(\therefore\) \(\boxed{\phantom{8}}\) \(\times\) DN = 15 \(\times\) 4
\(\therefore\) DN = \(\frac{\boxed{\phantom{60}}}{8}\)
\(\therefore\) DN = \(\boxed{\phantom{7.5}}\)
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