140559 Two light springs of force constants $k_{1}$ and $k_{2}$ and a block of mass $m$ are in one line $A B$ on a smooth horizontal table, such that one end of each spring is fixed to rigid support and other end is attached to block of mass $\mathrm{m} \mathrm{kg}$ as shown in figure. The frequency of vibration is:

140560 Two simple pendulums first (A) of bob mass $M_{1}$ and length $L_{1}$, second (B) of bob mass $M_{2}$ and length $L_{2} \cdot M_{1}=M_{2}$ and $L_{1}=2 L_{2}$. If the vibrational energy of both is same. Then which of the following is correct?

140561 The length of seconds pendulum is $1 \mathrm{~m}$ on the earth. If the mass and diameter of the planet is double than that of the earth, then the length of the seconds pendulum on the planet will be

140562 The bob of a simple pendulum performs S.H.M. with period ' $T$ ' in air and with period ' $T_{1}$ ' in water. Relation between ' $T$ ' and ' $T_{1}$ ' is (neglect friction due to water, density of the material of the bob is $=\frac{9}{8} \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$, density of water $=1 \mathrm{~g} / \mathrm{cc}$ )

140559 Two light springs of force constants $k_{1}$ and $k_{2}$ and a block of mass $m$ are in one line $A B$ on a smooth horizontal table, such that one end of each spring is fixed to rigid support and other end is attached to block of mass $\mathrm{m} \mathrm{kg}$ as shown in figure. The frequency of vibration is:

140560 Two simple pendulums first (A) of bob mass $M_{1}$ and length $L_{1}$, second (B) of bob mass $M_{2}$ and length $L_{2} \cdot M_{1}=M_{2}$ and $L_{1}=2 L_{2}$. If the vibrational energy of both is same. Then which of the following is correct?

140561 The length of seconds pendulum is $1 \mathrm{~m}$ on the earth. If the mass and diameter of the planet is double than that of the earth, then the length of the seconds pendulum on the planet will be

140562 The bob of a simple pendulum performs S.H.M. with period ' $T$ ' in air and with period ' $T_{1}$ ' in water. Relation between ' $T$ ' and ' $T_{1}$ ' is (neglect friction due to water, density of the material of the bob is $=\frac{9}{8} \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$, density of water $=1 \mathrm{~g} / \mathrm{cc}$ )

140559 Two light springs of force constants $k_{1}$ and $k_{2}$ and a block of mass $m$ are in one line $A B$ on a smooth horizontal table, such that one end of each spring is fixed to rigid support and other end is attached to block of mass $\mathrm{m} \mathrm{kg}$ as shown in figure. The frequency of vibration is:

140560 Two simple pendulums first (A) of bob mass $M_{1}$ and length $L_{1}$, second (B) of bob mass $M_{2}$ and length $L_{2} \cdot M_{1}=M_{2}$ and $L_{1}=2 L_{2}$. If the vibrational energy of both is same. Then which of the following is correct?

140561 The length of seconds pendulum is $1 \mathrm{~m}$ on the earth. If the mass and diameter of the planet is double than that of the earth, then the length of the seconds pendulum on the planet will be

140562 The bob of a simple pendulum performs S.H.M. with period ' $T$ ' in air and with period ' $T_{1}$ ' in water. Relation between ' $T$ ' and ' $T_{1}$ ' is (neglect friction due to water, density of the material of the bob is $=\frac{9}{8} \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$, density of water $=1 \mathrm{~g} / \mathrm{cc}$ )

140559 Two light springs of force constants $k_{1}$ and $k_{2}$ and a block of mass $m$ are in one line $A B$ on a smooth horizontal table, such that one end of each spring is fixed to rigid support and other end is attached to block of mass $\mathrm{m} \mathrm{kg}$ as shown in figure. The frequency of vibration is:

140560 Two simple pendulums first (A) of bob mass $M_{1}$ and length $L_{1}$, second (B) of bob mass $M_{2}$ and length $L_{2} \cdot M_{1}=M_{2}$ and $L_{1}=2 L_{2}$. If the vibrational energy of both is same. Then which of the following is correct?

140561 The length of seconds pendulum is $1 \mathrm{~m}$ on the earth. If the mass and diameter of the planet is double than that of the earth, then the length of the seconds pendulum on the planet will be

140562 The bob of a simple pendulum performs S.H.M. with period ' $T$ ' in air and with period ' $T_{1}$ ' in water. Relation between ' $T$ ' and ' $T_{1}$ ' is (neglect friction due to water, density of the material of the bob is $=\frac{9}{8} \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$, density of water $=1 \mathrm{~g} / \mathrm{cc}$ )