A 1 m long rod is half dipped in a swimming pool- The sunlight is incident at 45 - on the rod- Find the length of the shadow on the bed of swimming pool-

The correct option is D 81.5 cm Length of shadow =x+0.5m x=0.5 tan r sin r =.707× 3 4 or r= 32 ^∘ #160; 12 ^ ' Thus l=50 ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Physics

Total Internal Reflection

A 1 m long ...

Question

A $1 m$ long rod is half dipped in a swimming pool. The sunlight is incident at $45^{\circ}$ on the rod. Find the length of the shadow on the bed of swimming pool.

73 c m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

78.25 c m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

74.17 c m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

81.5 c m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $81.5 c m$
Length of shadow $= x + 0.5 m$

$x = 0.5 t a n r$

$sin r = .707 \times \frac{3}{4} o r$
$r = 32^{\circ} 12^{^{'}}$
$T h u s l = 50 (tan 32^{\circ} 12^{^{'}}) + 50 c m$
$= 50 (.6297) + 50 = 81.5 c m$

Suggest Corrections

Similar questions

Q. A pole of length 1.00 m stands half dipped in a swimming pool with water level 50.0 cm higher than the bed. The refractive index of water is 1.33 and sunlight is coming at an angle of 45° with the vertical. Find the length of the shadow of the pole on the bed.

Q. A swimming pool of length

= 25 m,

breadth

= 15 m

at height

= 2 m

is three fourths full of water. Find the volume of water in the pool.

Q. Sanjay dives into a swimming pool of length 135 m and breadth 14,000 cm.

Find the perimeter of the swimming pool.

The perimeter of a rectangular swimming pool is 64m. If the length and breadth are in the ratio of 5: 3. Find the length of the rectangular swimming pool in (m).

A swimming pool is 18m long, 8m wide and 2m deep. The capacity of the pool is 288 $m^{3}$

Join BYJU'S Learning Program

A 1 m long rod is half dipped in a swimming pool. The sunlight is incident at 45∘ on the rod. Find the length of the shadow on the bed of swimming pool.

The correct option is D 81.5 cmLength of shadow=x+0.5mx=0.5tanrsinr=.707×34orr=32∘12′Thusl=50(tan32∘12′)+50cm=50(.6297)+50=81.5cm

A $1 m$ long rod is half dipped in a swimming pool. The sunlight is incident at $45^{\circ}$ on the rod. Find the length of the shadow on the bed of swimming pool.

The correct option is D $81.5 c m$
Length of shadow $= x + 0.5 m$

$x = 0.5 t a n r$

$sin r = .707 \times \frac{3}{4} o r$
$r = 32^{\circ} 12^{^{'}}$
$T h u s l = 50 (tan 32^{\circ} 12^{^{'}}) + 50 c m$
$= 50 (.6297) + 50 = 81.5 c m$