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natima [27]
3 years ago
6

Kaushik had three pendulum bobs made of steel of the same length but different masses of 10 g, 50 g and 100 g respectively. When

he performed an experiment with them to find the time period, which of these statements would he consider to be correct? a. The time period is maximum with 10 g bob. b. The time period is maximum with 50 g bob. c. Time period remains the same in all. d. None of these is true.
Physics
1 answer:
Alexandra [31]3 years ago
8 0

Answer:

c. Time period remains the same in all.

Explanation:

In order to answer this question, we need to analyze the parameters, upon which the time period of a pendulum depends. We know that the time of a pendulum is given by the following formula:

T = 2π√(L/g)

where,

T = Time period

L = Length of pendulum

g = acceleration due to gravity

The formula clearly shows that the time period of the pendulum depends only upon the length of pendulum and value of g. And the time period of a pendulum does not depend upon the mass of the bob. Hence, the time period for each of the three pendulums will remain same. So, the correct option will be:

<u>c. Time period remains the same in all.</u>

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A 19.12 g mixture of Ca(NO3)2 and KCl is dissolved in 149 g of water. The freezing point of the solution was measured as −5.77 ∘
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Answer:

The mass percentage of calcium nitrate is 31.23%.

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Let the the mass of calcium nitrate be x and mass of potassium chloride be y.

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x + y = 19.12 g..(1)

Mass of solvent = 149 g = 0.149 kg

Freezing point of the solution,T_f = -5.77 °C

Molal freezing constant of water = 1.86 °C/m =1.86 °C/(mol/kg)

The van't Hoff factor contribution by calcium nitrate is 3 and by potassium chloride is 2.So:

i = 3

i' = 2

Freezing point of water = T = 0°C

\Delta T_f=T-T_f=0^oC-(-5.77^oC)=5.77^oC

\Delta T_f=i\times K_f\times m

Molality=m(mol/kg)=\frac{\text{Moles of solute}}{\text{mass of solvent in kg}}

5.77^oC=1.86 ^oC/(mol/kg)\times (\frac{ i\times x}{164 g/mol\times 0.149 kg}+\frac{i'\times y}{74.5 g/mol0.149 kg})

On solving we get:

\frac{3x}{164 g/mol}+\frac{2x}{74.5 g/mol}=0.4622 mol....(2)

Solving equation (1)(2) for x and y:

x =5.973 g

y = 13.147 g

Mass percent of Ca(NO_3)_2 in the mixture:

\frac{x}{19.21 g}\times 100=\frac{5.973 g}{19.12 g}=31.23\%

The mass percentage of calcium nitrate is 31.23%.

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