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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Physics Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Three Motion in a Straight Line

When a ball is dropped into a lake form a height 4.9 m above the water level, it hits the water with a velocity v and then sinks to the bottom with the constant velocity v. It reaches the bottom of the lake 4.0 s after it is dropped. The approximate depth of the lake is

Option 1 -

19.6 m

Option 2 -

29.4 m

Option 3 -

39.2 m

Option 4 -

73.5 m

4 Views | Posted 3 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

3 months ago

Correct Option - 2

Detailed Solution:

Total time (T) = 4 sec

Given u = 0 m/sec

a = g = 9.8 m/sec²

h = 4.9 m, t =?

$\Rightarrow h = u t + \frac{1}{2} a t^{2}$

$4.9 = 0 . t + \frac{1}{2} \times 9.8 t^{2}$

t = 1 sec

‘v’ be the velocity with which ball hits the water v = u + at

= 0 + 9.8 × 1 = 9.8 m/sec

Time taken to reach the bottom of the lake from surface of the lake

= 4 – 1 = 3 sec

v = 9.8 m/sec

$H = u t + \frac{1}{2} g t^{2} = 9.8 \times 3 + \frac{1}{2} \times 9.8 \times 9$

29.4 + 4.9 × 9 = 29.4 + 44.1

H = 73.5 m

Total time (T) = 4 secGiven u = 0 m/seca = g = 9.8 m/sec2h = 4.9 m, t =? <math> <mrow> <mo>⇒</mo> <mi>h</mi> <mo>=</mo> <mi>u</mi> <mi>t</mi> <mo>+</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mi>a</mi> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math>              <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1813494210"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mn>4</mn> <mo>.</mo> <mn>9</mn> <mo>=</mo> <mn>0</mn> <mo>.</mo> <mi>t</mi> <mo>+</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>×</mo> <mn>9</mn> <mo>.</mo> <mn>8</mn> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math>              <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1813494211"> </o:OLEObject></xml><! [endif]-> t = 1 sec‘v’ be the velocity with which ball hits the water v = u + at= 0 + 9.8 × 1 = 9.8 m/secTime taken to reach the bottom of the lake from surface of the lake= 4 – 1 = 3 secv = 9.8 m/sec <math> <mrow> <mi>H</mi> <mo>=</mo> <mi>u</mi> <mi>t</mi> <mo>+</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mi>g</mi> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <mn>9</mn> <mo>.</mo> <mn>8</mn> <mo>×</mo> <mn>3</mn> <mo>+</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>×</mo> <mn>9</mn> <mo>.</mo> <mn>8</mn> <mo>×</mo> <mn>9</mn> </mrow> </math>              <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1813494212"> </o:OLEObject></xml><! [endif]-> 29.4 + 4.9 × 9 = 29.4 + 44.1H = 73.5 m<div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1751968708phpitlVWj.jpeg" alt="" width="247" height="251" loading="lazy"></picture></div></div>

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Let 't' be the time taken by B to meat A.

for A, u = 0

t = [2 + t]

h = 80m

a = 10 m/sec²

h = ut + $\frac{1}{2} a t^{2}$

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Let ‘t’ be the time taken by B to meat

for A, u = 0

t = [2 + t]

h = 80m

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When a ball is dropped into a lake form a height 4.9 m above the water level, it hits the water with a velocity v and then sinks to the bottom with the constant velocity v. It reaches the bottom of the lake 4.0 s after it is dropped. The approximate depth of the lake is

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