The given equation is:
\(\frac{mgh}{4} = mL\)
Where: - \( m \) is the mass, - \( g \) is the acceleration due to gravity, - \( h \) is the height, - \( L \) is a given length.
Rearranging the equation to solve for \( h \):
\(h = \frac{4L}{g}\)
Substituting the given values into the formula:
\(h = \frac{4 \times 3.4 \times 10^{5}}{10}\)
Simplifying:
\(h = 136 \, \text{km}\)
The calculated height \( h \) is \( 136 \, \text{km} \).
So, here In the question it is mentioned that an ice piece falls down from a height ‘h’ and it is melted completely.
During the fall, only \(\frac{1}{4}\)th of the heat is absorbed by the ice and the rest remaining energy is converted into heat.
So, Here the Formula used is: mgh=mL.
We are already given the latent heat of ice and the value of g.
Which are as follows: Latent heat of ice =3.4×15J/k
g = 10N/kg
Now, we will consider ‘m’ to be the mass of ice.
So, when ice falls down from the height ‘h’, we know that its potential energy is converted into heat energy.
Which means gravitational potential energy is equivalent to the latent heat of ice.
As the mass of ice is ‘m’ here, Latent heat can be written as follows: ′mL′,
Where ‘L’ belongs to the latent heat of ice.
Hence, we can say that, mgh=mL, Where the ‘mgh′ is the potential energy and ‘mL′ is the latent heat.
As mentioned in the question, only one-fourth of the heat is absorbed by the ice during fall.
Therefore we can say that, \(\frac{mgh}{4}\)=mL
After, solving the above equation, we get ‘h’ as, h=\(\frac{4L}{g}\)
After Substituting the values of ‘L’ and ‘g’ in the above equation, we get \(h=\frac{4\times3.4\times10^{5}/kg}{10N/kg}\)
\(h=136\times10^{3}m\)
\(h=136km\)
As a result, the ice falls from a height h=136 km.
So, the correct answer is “Option B”.
A metal plate of area 10-2m2 rests on a layer of castor oil, 2 × 10-3m thick, whose viscosity coefficient is 1.55 Ns/m2. The approximate horizontal force required to move the plate with a uniform speed of 3 × 10-2ms-1 is:
In a Vernier caliper, \(N+1\) divisions of vernier scale coincide with \(N\) divisions of main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is: