This problem requires applying Kepler's Third Law of Planetary Motion, which relates a planet's orbital period ($T$) to the semi-major axis ($R$) of its orbit around the Sun. The law states that the square of the period is proportional to the cube of the semi-major axis, or $\frac{T^2}{R^3} = \text{constant}$.
We can compare the Earth and planet X:
According to Kepler's Third Law:
$ \frac{T_E^2}{R_E^3} = \frac{T_X^2}{R_X^3} $
We are given:
We need to find $R_X$. Rearranging the formula to solve for $R_X^3$: $ R_X^3 = R_E^3 \times \left(\frac{T_X}{T_E}\right)^2 $
Substitute the values:
$ R_X^3 = (1.5 \times 10^{11} \ m)^3 \times \left(\frac{8 \ years}{1 \ year}\right)^2 $
$ R_X^3 = (1.5^3 \times (10^{11})^3) \times 8^2 \ m^3 $
$ R_X^3 = (3.375 \times 10^{33}) \times 64 \ m^3 $
$ R_X^3 = 216 \times 10^{33} \ m^3 $
Now, take the cube root to find $R_X$: $ R_X = \sqrt[3]{216 \times 10^{33}} \ m $
$ R_X = \sqrt[3]{216} \times \sqrt[3]{10^{33}} \ m $
$ R_X = 6 \times 10^{11} \ m $
Therefore, the distance of planet X from the Sun is $6 \times 10^{11}$ m.
A uniform bar of length 12 cm and mass $20m$ lies on a smooth horizontal table. Two point masses $m$ and $2m$ are moving in opposite directions with same speed of $v$ and in the same plane as the bar, as shown in figure. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency $\omega$. The ratio of $v$ and $\omega$ is :
Match the LIST-I with LIST-II
| List-I: | List-II: |
| A. Magnetic induction | I. $[M L T^{-2} A^{-2}]$ |
| B. Magnetic flux | II. $[M L^2 T^{-2} A^{-2}]$ |
| C. Magnetic permeability | III. $[M L^0 T^{-2} A^{-1}]$ |
| D. Self inductance | IV. $[M L^2 T^{-2} A^{-1}]$ |
Choose the correct answer from the options given below:
In the given figure the blocks $A$, $B$ and $C$ weigh 4 kg, 6 kg and 8 kg respectively. The co-efficient of sliding friction between any two surfaces is 0.5. The force $\vec{F}$ required to slide the block $C$ with constant speed is ______ N. (Use $g = 10 \text{ m/s}^2$)

In case of vertical circular motion of a particle by a thread of length $r$ if the tension in the thread is zero at an angle $30^\circ$ shown in figure, the velocity at the bottom point ($A$) of the circular path is
($g = \text{gravitational acceleration}$)

A uniform bar of length 12 cm and mass $20m$ lies on a smooth horizontal table. Two point masses $m$ and $2m$ are moving in opposite directions with same speed of $v$ and in the same plane as the bar, as shown in figure. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency $\omega$. The ratio of $v$ and $\omega$ is :
Match the LIST-I with LIST-II
| List-I: | List-II: |
| A. Magnetic induction | I. $[M L T^{-2} A^{-2}]$ |
| B. Magnetic flux | II. $[M L^2 T^{-2} A^{-2}]$ |
| C. Magnetic permeability | III. $[M L^0 T^{-2} A^{-1}]$ |
| D. Self inductance | IV. $[M L^2 T^{-2} A^{-1}]$ |
Choose the correct answer from the options given below: