SSC 10th Science Maharashtra Board: Numerical Problem 4
Chapter 1 (Gravitation) - Question 5 (d)
Note: This is the 4th numerical problem listed in the Gravitation chapter exercise.
An object thrown vertically upwards reaches a height of $500\text{ m}$. What was its initial velocity? How long will the object take to come back to the earth? Assume $g = 10\text{ m/s}^2$.
Solution:
Given:
Maximum height ($s$ or $h$) = $500\text{ m}$
Final velocity at maximum height ($v$) = $0\text{ m/s}$
Acceleration due to gravity ($a$) = $-g = -10\text{ m/s}^2$ (negative because the object is moving against gravity)To find:
Initial velocity ($u$) = ?
Total time taken to return to earth ($t_{total}$) = ?Formulae:
1) $v^2 = u^2 + 2as$
2) $v = u + at$
3) Total time $t_{total} = t_{up} + t_{down}$Calculation:
Step 1: Find initial velocity ($u$)
Using Newton's third equation of motion:
$$0^2 = u^2 + 2(-10)(500)$$
$$0 = u^2 - 10000$$
$$u^2 = 10000$$
$$u = 100\text{ m/s}$$Step 2: Find time taken to go up ($t_{up}$)
Using Newton's first equation of motion:
$$v = u + at_{up}$$
$$0 = 100 + (-10)t_{up}$$
$$10t_{up} = 100$$
$$t_{up} = 10\text{ s}$$Step 3: Total time to come back to the earth
The time taken to reach the maximum height is equal to the time taken to fall back to the ground (ignoring air resistance).
$$t_{down} = t_{up} = 10\text{ s}$$
$$t_{total} = 10\text{ s} + 10\text{ s} = 20\text{ s}$$Answer:
The initial velocity of the object was $100\text{ m/s}$ and it will take $20\text{ s}$ to come back to the earth.
Chapter 4 (Effects of Electric Current) - Question 4
Note: If you were referring to Question 4 from the Effects of Electric Current chapter (which is also a numerical problem), the solution is below.
A cell is connected to a $9\text{ }\Omega$ resistance, because of which heat of $400\text{ J}$ is produced per second due to current flowing through it. Obtain the potential difference applied across the resistance.
Solution:
Given:
Resistance ($R$) = $9\text{ }\Omega$
Heat produced per second ($P$) = $\frac{H}{t} = 400\text{ J/s} = 400\text{ W}$To find:
Potential difference ($V$) = ?Formula:
Power $P = \frac{V^2}{R}$Calculation:
$$400 = \frac{V^2}{9}$$
$$V^2 = 400 \times 9$$
$$V^2 = 3600$$
$$V = \sqrt{3600}$$
$$V = 60\text{ V}$$Answer:
The potential difference applied across the resistance is $60\text{ V}$.
