Answer
Verified
91.2k+ views
Hint Here we know the radius value of mercury and change so we find the large drop and small drop by using the formula ratio of surface density change according to some change in the formula for calculating the ratio than the volume of the small and large drops in the problem.
Useful formula:
The surface charge density formula is given by,
\[\sigma \; = \;\dfrac{q}{A}{\text{ }}\]
Where,
\[\sigma \;\]is surface charge density
$q$ is charge
$A$ is surface area
Complete step by step answer
Given by,
Let the radius of big drop $ = R$
Volume of the big drop $ = \dfrac{4}{3}\pi {R^3}$
Volume of a small drop $ = \dfrac{4}{3}\pi {r^3}$
Volume of $64$ small drops $ = 64 \times \dfrac{4}{3}\pi {r^3}$
Here the above equation is equal,
Volume of the big drop is equal to Volume of $64$ small drops
We get,
\[\dfrac{4}{3}\pi {r^3} = 64 \times \dfrac{4}{3}\pi {r^3}\]
Common factor is canceled,
Here,
${R^3} = 64{r^3}$
On simplifying,
$R = 4r$
Charge on a small drop $ = q$
Charge on a big drop$ = Q$
The value of $Q = 64q$
Small drop ${\sigma _1} = \dfrac{q}{{4\pi {r^2}}}$
Big drop \[{\sigma _2} = \dfrac{Q}{{4\pi {R^2}}}\]
Substituting the given value in above equation,
We get,
\[{\sigma _2} = \dfrac{{64q}}{{4\pi {{(4r)}^2}}}\]
On simplifying,
\[{\sigma _2} = \dfrac{{64q}}{{64\pi {r^2}}}\]
Common factor canceled,
\[{\sigma _2} = \dfrac{q}{{\pi {r^2}}}\]
According to the surface density,
Here,
\[\dfrac{{{\sigma _1}}}{{{\sigma _2}}} = \dfrac{q}{{4\pi {r^2}}} \times \dfrac{q}{{\pi {r^2}}}\]
Again, the common factors are canceled,
We get,
\[\dfrac{{{\sigma _1}}}{{{\sigma _2}}} = \dfrac{1}{4}\]
Hence,
\[{\sigma _1}:{\sigma _2} = 1:4\]
Thus, The option B is correct answer
Note Charging density is known, according to electromagnetism, as a measure of the electrical charge per unit volume of space in one, two or three dimensions. The linear surface or volume charge density is the quantity of electric charge per surface area or volume, to be specific.
Useful formula:
The surface charge density formula is given by,
\[\sigma \; = \;\dfrac{q}{A}{\text{ }}\]
Where,
\[\sigma \;\]is surface charge density
$q$ is charge
$A$ is surface area
Complete step by step answer
Given by,
Let the radius of big drop $ = R$
Volume of the big drop $ = \dfrac{4}{3}\pi {R^3}$
Volume of a small drop $ = \dfrac{4}{3}\pi {r^3}$
Volume of $64$ small drops $ = 64 \times \dfrac{4}{3}\pi {r^3}$
Here the above equation is equal,
Volume of the big drop is equal to Volume of $64$ small drops
We get,
\[\dfrac{4}{3}\pi {r^3} = 64 \times \dfrac{4}{3}\pi {r^3}\]
Common factor is canceled,
Here,
${R^3} = 64{r^3}$
On simplifying,
$R = 4r$
Charge on a small drop $ = q$
Charge on a big drop$ = Q$
The value of $Q = 64q$
Small drop ${\sigma _1} = \dfrac{q}{{4\pi {r^2}}}$
Big drop \[{\sigma _2} = \dfrac{Q}{{4\pi {R^2}}}\]
Substituting the given value in above equation,
We get,
\[{\sigma _2} = \dfrac{{64q}}{{4\pi {{(4r)}^2}}}\]
On simplifying,
\[{\sigma _2} = \dfrac{{64q}}{{64\pi {r^2}}}\]
Common factor canceled,
\[{\sigma _2} = \dfrac{q}{{\pi {r^2}}}\]
According to the surface density,
Here,
\[\dfrac{{{\sigma _1}}}{{{\sigma _2}}} = \dfrac{q}{{4\pi {r^2}}} \times \dfrac{q}{{\pi {r^2}}}\]
Again, the common factors are canceled,
We get,
\[\dfrac{{{\sigma _1}}}{{{\sigma _2}}} = \dfrac{1}{4}\]
Hence,
\[{\sigma _1}:{\sigma _2} = 1:4\]
Thus, The option B is correct answer
Note Charging density is known, according to electromagnetism, as a measure of the electrical charge per unit volume of space in one, two or three dimensions. The linear surface or volume charge density is the quantity of electric charge per surface area or volume, to be specific.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
Electric field due to uniformly charged sphere class 12 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Derive an expression for maximum speed of a car on class 11 physics JEE_Main
3 mole of gas X and 2 moles of gas Y enters from the class 11 physics JEE_Main
A soldier with a machine gun falling from an airplane class 11 physics JEE_MAIN
Velocity of car at t 0 is u moves with a constant acceleration class 11 physics JEE_Main