A simple pendulum has time period ‘$\mathbf{T}$’ depending on length ‘$\mathbf{l}$’ and gravitational acceleration ‘$\mathbf{g}$’. Using dimensional analysis, the form of ‘$\mathbf{T}$’ is:

If $\mathbf{T \propto}\mathbf{l}^{\mathbf{a\ }}\mathbf{g}^{\mathbf{b}}$, then the value of ‘$\mathbf{a}$’ for a simple pendulum is:

The time period of a pendulum is independent of:

Escape velocity ‘$\mathbf{v}$’ depends on gravitational acceleration ‘$\mathbf{g}$’ and radius ‘$\mathbf{r}$’ of a planet. The correct dimensional formula is:

The dimensions of $\mathbf{gr}$ are:

The speed of waves on a stretched string depends upon tension ‘$\mathbf{F}$’ and mass per unit length ‘$\mathbf{\mu}$’. The correct relation is:

If $\mathbf{v \propto}\mathbf{F}^{\mathbf{a\ }}\mathbf{\mu}^{\mathbf{b}}$, the value of ‘$\mathbf{a}$’ is:

The dimensions of mass per unit length are:

The kinetic energy ‘$\mathbf{E}$’ of a body depends on mass ‘$\mathbf{m}$’ and velocity ‘$\mathbf{v}$’. Dimensional analysis predicts:

Dimensional analysis cannot determine:

The force ‘$\mathbf{F}$’ acting on a body depends on mass ‘$\mathbf{m}$’ and acceleration ‘$\mathbf{a}$’. Dimensional analysis gives:

In dimensional analysis, if the number of unknown exponents exceeds the number of fundamental dimensions, then:

If pressure ‘$\mathbf{P}$’ depends on force ‘$\mathbf{F}$’ and area ‘$\mathbf{A}$’, then:

Which quantity is commonly expressed as $\mathbf{v \propto}\sqrt{\mathbf{gr}}\mathbf{\ }$?

If a formula derived dimensionally is $\mathbf{Q \propto}\mathbf{a}^{\mathbf{2}}\mathbf{\ }\mathbf{b}^{\mathbf{- 1}}$, then increasing $\mathbf{a}$ by $\mathbf{2}$ times changes $\mathbf{Q}$ by:

The dimensions of energy are:

A student derives $\mathbf{T = k}$ $\sqrt{\frac{\mathbf{l}}{\mathbf{g}}}$. The symbol $\mathbf{k}$ represents:

The speed of a wave depends on tension ‘$\mathbf{F}$’ and linear density ‘$\mathbf{\mu}$’. If tension becomes four times, speed becomes:

A formula derived dimensionally is most useful for:

If gravitational acceleration becomes four times, the period of a pendulum becomes:

Which quantity has dimensions $\mathbf{\lbrack}\mathbf{M\ L}^{\mathbf{- 1}}\mathbf{\rbrack}$?

If length of a pendulum becomes four times, its period becomes:

Which of the following can be derived using dimensional analysis?

If escape velocity $\mathbf{v \propto}\sqrt{\mathbf{gr}}$, doubling radius while keeping$\mathbf{g}$ constant makes $\mathbf{v}$:

If $\mathbf{Q \propto}\mathbf{a}^{\mathbf{x}}\mathbf{\ }\mathbf{b}^{\mathbf{y}}$, dimensions are compared to:

Why is dimensional analysis called a powerful tool?

Which step comes first in dimensional derivation?

Which formula cannot be completely obtained by dimensional analysis?

The dimensions of $\mathbf{F\ /\ \mu}$ are:

Taking square root of $\mathbf{F\ /\ \mu}$ gives dimensions of:

Which of the following is a limitation of dimensional analysis?

If kinetic energy depends on mass and velocity only, dimensional analysis predicts the power of velocity to be:

The dimensions of $\mathbf{g\ /\ }\mathbf{l}$ are:

The square root of $\mathbf{g\ /\ }\mathbf{l}$ has dimensions:

A correct dimensional derivation requires: