\(5.74\) g of a substance occupies \(1.2~\text{cm}^3\). Its density by keeping the significant figures in view is:
1. \(4.7333~\text{g/cm}^3\)
2. \(3.8~\text{g/cm}^3\)
3. \(4.8~\text{g/cm}^3\)
4. \(3.7833~\text{g/cm}^3\)

 The SI unit of energy is \(\mathrm{J = \text{kg}\left(m\right)^{2} s^{- 2}}\); that of speed \(v\) is \(\text{ms}^{- 1}\) and of acceleration \(a\) is \(\text{ms}^{- 2}\). Which of the formula for kinetic energy (\(K\)) given below can you rule out on the basis of dimensional arguments (m stands for the mass of the body)?
 

Choose the correct option:

Let us consider an equation \(\dfrac{1}{2}mv^2=mgh\) where \(m\) is the mass of the body, \(v\) its velocity, \(g\) is the acceleration due to gravity and \(h\) is the height. The equation is:

Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length \((l)\), the mass of the bob \((m)\) and acceleration due to gravity \((g)\). Expression for its time period is:

1.  \(T   =   \dfrac{1}{2 \pi} \sqrt{\dfrac{l}{g}} \)

2.  \(T   =   2 \pi \left(\dfrac{l}{g}\right) \)

3.  \(T   =   2 \pi \sqrt{\dfrac{l}{g}} \)

4.  \(T   =   2 \pi \sqrt{\dfrac{g}{l}} \)

If force \([F]\), acceleration \([A]\) and time \([T]\) are chosen as the fundamental physical quantities, then find the dimensions of energy:

Each side of a cube is measured to be \(7.203~\text{m}\). What are the total surface area and the volume respectively of the cube to appropriate significant figures? 

The period of oscillation of a simple pendulum is $\mathrm{T}<mspace\; width="1em"></mspace>=2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. The percentage error in g is:

1.  $2\%$

2.  $3\%$

3.  $5\%$

4.  $0\%$

The relative error in \(Z,\) if \(Z=\frac{A^{4}B^{1/3}}{CD^{3/2}}\) is:
1. \(\frac{\Delta A}{A}+\frac{\Delta B}{B}+\frac{\Delta C}{C}+\frac{\Delta D}{D}\)

2. \(4\frac{\Delta A}{A}+\frac{1}{3}\frac{\Delta B}{B}-\frac{\Delta C}{C}- \frac{3}{2}\frac{\Delta D}{D}\)

3. \(4\frac{\Delta A}{A}+\frac{1}{3}\frac{\Delta B}{B}+\frac{\Delta C}{C}+\frac{2}{3}\frac{\Delta D}{D}\)

4. \(4\frac{\Delta A}{A}+\frac{1}{3}\frac{\Delta B}{B}+\frac{\Delta C}{C}+\frac{3}{2}\frac{\Delta D}{D}\)

Two resistors of resistances ${\mathrm{R}}_1<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\left(100<mspace\; width="1em"></mspace>\pm <mspace\; width="1em"></mspace>3\right)$ ohm and ${\mathrm{R}}_2<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\left(200<mspace\; width="1em"></mspace>\pm <mspace\; width="1em"></mspace>4\right)$ ohm are connected in parallel. The equivalent resistance of the parallel combination is:

1.  (300 ± 7) ohm

2.  (66.7 ± 7) ohm

3.  (66.7 ± 1.8) ohm

4.  (100 ± 1) ohm