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Varvara68 [4.7K]

3 years ago

6

The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actu

al depth of the liquid in the tank?

1 answer:

Juliette [100K]3 years ago

3 0

$\orange{\underline{\huge{\bold{\textit{\green{\bf{QUESTION}}}}}}}$

The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actual depth of the liquid in the tank?

${\bold{\blue{GIVEN}}}$

REFRACTIVE INDEX = 1.3

APPARENT DEPTH = 7.7 cm

${ \bold{\green{To \: Find}}}$

REAL DEPTH OF THE OBJECT.

${\red{FORMULA \: \: USED }}$

$Reflective \: Index = \frac{Real \: Depth}{Apparent \: Depth }$

$\huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}$

Refractive Index = 1.3

Apparent Depth = 7.7 cm

Putting the values in the formula:-

$Reflective \: Index = \frac{Real \: Depth}{Apparent \: Depth } \\ \\ 1.3 = \frac{Real \: Depth}{7.7 \: cm} \\ \\ 1.3 \times 7.7 = Real \: Depth \\ \\ 10.01 \: \: cm = Real \: Depth$

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Answer:

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Explanation:

Given,

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$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

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From the conservation of linear momentum,

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For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

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Given,

Time of impact = t = $25\times 10^{-3}\ sec$

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