Question

plz solve physics class 10​

Answer 1

Answer:

the answer of question 'b' is 40 ohm

Answer 2

Question 1: An electric refrigerator rated 400W operates 10 hours/day. What is the cost of the energy required to operate it for 30 days at Rs. 4.00 per kWH?

Solution:

Power = 400W

Time taken per day = 10 hours.

ATQ,

We have to find the Energy required to operate the refrigerator for 30 days. Since in one day it operates for 10 hours, In 30 days, it operates for:

Time taken in 30 days = 30 × 10

∴ Time taken in 30 days = 300 hours.

Since the cost is to be found in terms of kWH (Kilo Watt Hour), we'll 400w in terms of kW.

We know that to convert Watt(W) to Kilo Watt (kW), we divide the value of Watt by thousand.

P = 400W

$\sf P = \dfrac{400}{1000}$

P = 0.4 kW.

Now, the energy required to operate the refrigerator will be,

E = P × T

E = 0.4 × 300

E = 120 kWH.

Now, we multiply the Energy by 4.00 Rs to get the total cost.

Cost of energy to operate the Refrigerator = Cost per kWH × Energy.

Cost required = 4.00 × 120

Cost required = 480 Rs.

Hence, the cost of the energy required to operate the refrigerator for 30 days at the cost of 4Rs is 480 Rs.

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Question 2: A wire of given material having length 'l' and area of cross-section 'A' has a resistance of 10 Ω. What would be the resistance of another wire of the same material having length 2l and area of cross-section A/2?

Solution:

Original Wire:

Resistivity = ρ.

Length = l.

Area of cross section = A.

Resistance = 10 Ω.

We know that,

$\Longrightarrow \ \sf R = \rho \times \dfrac{l}{A}$

$\sf \Longrightarrow \ 10 = \ \rho \times \dfrac{l}{A}$

Let the above be Equation 1.

New Wire:

Resistivity = ρ.

Length = 2l.

Area of cross section = A/2

Resistance = ? Ω.

$\sf \Longrightarrow \ R = \ \rho \times \dfrac{l}{A}$

$\sf \Longrightarrow \ R = \ \rho \times \dfrac{2l}{A/2}$

$\sf \Longrightarrow \ R = \ \rho \times \dfrac{4l}{A}$

$\sf \Longrightarrow \ R = \ 4 \times \rho\dfrac{l}{A}$

Substitute Equation 1 above.

$\sf \Longrightarrow \ R = \ 4 \times 10$

$\sf \Longrightarrow \ R = \ 40 \Omega$

Answer: 40 Ω.