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. A squirrel climbs a Cylindrical tree in 20 seconds in such a way that it makes exactly 4 complete turns round the tree while it touches the bottom and top ends of the tree. If the Diameter and Height of the tree is 26/11cm and 24 cm. Find the speed of the squirrel (in m/s) .?
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. A squirrel climbs a Cylindrical tree in 20 seconds in such a way tha...
**Problem Analysis**

To find the speed of the squirrel, we need to determine the distance it travels and the time it takes. We are given that the squirrel climbs the tree in 20 seconds and makes 4 complete turns while touching the bottom and top ends of the tree.

To calculate the distance, we need to find the circumference of the tree, as the squirrel travels this distance in each turn. We are given the diameter of the tree, which can be used to calculate the radius.

Once we have the distance and the time, we can calculate the speed using the formula:

Speed = Distance / Time

**Calculating the Circumference**

The diameter of the tree is given as 26/11 cm. To calculate the circumference, we need to find the radius.

The formula for the circumference of a circle is:

Circumference = 2 * π * radius

Given that the diameter is 26/11 cm, the radius can be calculated as:

Radius = diameter / 2 = (26/11) / 2 = 13/11 cm

Substituting the value of radius in the formula for circumference, we get:

Circumference = 2 * π * (13/11) cm

**Calculating the Distance**

Since the squirrel makes 4 complete turns around the tree, the distance it travels is 4 times the circumference of the tree. Therefore, the distance can be calculated as:

Distance = 4 * Circumference

Substituting the value of circumference, we get:

Distance = 4 * 2 * π * (13/11) cm

Simplifying the expression, we get:

Distance = 8 * π * (13/11) cm

**Calculating the Speed**

The speed of the squirrel is given by the formula:

Speed = Distance / Time

We are given that the squirrel climbs the tree in 20 seconds, so the time is 20 seconds. Substituting the values of distance and time, we get:

Speed = (8 * π * (13/11) cm) / 20 seconds

To convert the speed from cm/s to m/s, we need to divide the speed by 100. Therefore, the final speed in m/s is:

Speed = [(8 * π * (13/11) cm) / 20 seconds] / 100 m/s

Simplifying the expression, we get:

Speed = (8 * π * (13/11)) / (20 * 100) m/s

Speed = (104π/2200) m/s

Speed ≈ 0.0473 m/s

Therefore, the speed of the squirrel is approximately 0.0473 m/s.
Community Answer
. A squirrel climbs a Cylindrical tree in 20 seconds in such a way tha...
मुझे पूरा यकीन हीं है लेकिन 17/22cm/second but here ask. us m/s so tha answer is may be 17/2200 mete I can't upload photo of my solution I don't know why. may be some bugs...
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. A squirrel climbs a Cylindrical tree in 20 seconds in such a way that it makes exactly 4 complete turns round the tree while it touches the bottom and top ends of the tree. If the Diameter and Height of the tree is 26/11cm and 24 cm. Find the speed of the squirrel (in m/s) .?
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. A squirrel climbs a Cylindrical tree in 20 seconds in such a way that it makes exactly 4 complete turns round the tree while it touches the bottom and top ends of the tree. If the Diameter and Height of the tree is 26/11cm and 24 cm. Find the speed of the squirrel (in m/s) .? for SSC CGL 2024 is part of SSC CGL preparation. The Question and answers have been prepared according to the SSC CGL exam syllabus. Information about . A squirrel climbs a Cylindrical tree in 20 seconds in such a way that it makes exactly 4 complete turns round the tree while it touches the bottom and top ends of the tree. If the Diameter and Height of the tree is 26/11cm and 24 cm. Find the speed of the squirrel (in m/s) .? covers all topics & solutions for SSC CGL 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for . A squirrel climbs a Cylindrical tree in 20 seconds in such a way that it makes exactly 4 complete turns round the tree while it touches the bottom and top ends of the tree. If the Diameter and Height of the tree is 26/11cm and 24 cm. Find the speed of the squirrel (in m/s) .?.
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