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*Simulator*

**Previous Years AIEEE/JEE Mains Questions**

(1) One degree

(2) Half degree

(3) One minute

(4) Half minute

**[AIEEE – 2009]**

**Sol.**(3)

Least count of vernier callipers

L.C. = 1MSD – 1VSD

But here $29 \mathrm{MSD}=30 \mathrm{VSD} \Rightarrow 1 \mathrm{VSD}=\frac{29}{30} \mathrm{MSD}$

$\Rightarrow \mathrm{L.C.}=1 \mathrm{MSD}-\frac{29}{30} \mathrm{MSD}=$

$\frac{1}{30} \mathrm{MSD}=\frac{1}{30} \times 0.5^{\circ}=\left(\frac{1}{60}\right)^{\circ}=1$ minute.

(1) (ƒ, ƒ)

(2) (4ƒ, 4ƒ)

(3) (2ƒ, 2ƒ)

(4) $\left(\frac{f}{2}, \frac{f}{2}\right)$

**[AIEEE – 2009]**

**Sol.**(3)

(1) 4, 4, 2 (2) 5, 1, 2 (3) 5, 1, 5 (4) 5, 5, 2

**[AIEEE – 2010]**

**Sol.**(2)

5, 1, 2 resp.

Main scale reading : 0 mm.

Circular scale reading : 52 divisions

Given that 1 mm on main scale corresponds to 100 divisions of the circular scale.

The diameter of wire from the above date is :-

(1) 0.026 cm (2) 0.005 cm (3) 0.52 cm (4) 0.052 cm

**[AIEEE – 2011]**

**Sol.**(4)

Least count $=\frac{1 \mathrm{mm}}{100}=0.01 \mathrm{mm}$

Diameter of the wire = 0 + 52 × 0.01 mm = 0.52 mm = 0.052 cm

(1) 59 degree (2) 58.59 degree (3) 58.77 degree (4) 58.65 degree

**[AIEEE – 2012]**

**Sol.**(4)

Least count = MSD – VSD

where $\mathrm{MSD}=0.5^{\circ} \& 30 \mathrm{VSD}=29 \mathrm{MSD}$

so least count $=0.5^{\circ}-\left(\frac{29}{30}\right) \times 0.5^{\circ}=\left(\frac{0.5}{30}\right)^{\circ}$

Reading $=58.5^{\circ}+(09) \times\left(\frac{0.5}{30}\right)^{\circ}=58.65^{\circ}$

(1) 3% (2) 6% (3) zero (4) 1%

** [AIEEE – 2012]**

**Sol.**(2)

$\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}} \Rightarrow \frac{\Delta \mathrm{R}}{\mathrm{R}}=\pm\left(\frac{\Delta \mathrm{V}}{\mathrm{V}}+\frac{\Delta \mathrm{I}}{\mathrm{I}}\right)$

$=\pm(3+3) \%=\pm 6 \%$

(1) 0.5 mA (2) 0.05 mA (3) 0.2 mA (4) 0.02 mA

**[JEE-Mains 2014]**

**Sol.**(3)

Given current $\mathrm{I}=\left(e^{1000 \mathrm{V} / \mathrm{T}}-1\right) \mathrm{mA}$

$\Rightarrow \mathrm{I}+1=e^{1000 \mathrm{V} / \mathrm{T}}$

$\mathrm{d} \mathrm{l}=\frac{1000}{\mathrm{T}}\left[e^{\frac{1000 \mathrm{v}}{\mathrm{T}}}\right] \mathrm{d} \mathrm{V}$

$\mathrm{d} \mathrm{l}=\frac{1000}{\mathrm{T}}[1+1] \mathrm{dV}=\frac{1000}{300}[6] \times(0.01)$

dI = 0.2 mA

(1) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.

(2) A screw gauge having 50 divisions in the circular scale and pitch as 1 mm.

(3) A meter scale

(4) A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1 cm.

**[JEE-Mains 2014]**

**Sol.**(4)

Least count of varnier calliper is 0.01 cm

Hence it matches with the reading.

(1)1% (2) 5% (3) 2% (4) 3%

**[JEE-Mains 2015]**

**Sol.**(4)

$\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$

$\mathrm{g}=\frac{4 \pi^{2} \ell}{\mathrm{T}^{2}}$

$\frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{\Delta \ell}{\ell}+2 \frac{\Delta \mathrm{T}}{\mathrm{T}}$

$\frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{1 \times 10^{-3}}{20 \times 10^{-2}}+2 \times \frac{1}{100 \times \frac{90}{100}}$

$\therefore \frac{\Delta \mathrm{g}}{\mathrm{g}} \times 100=2.722 \% \approx 3 \%$

(1) 0.50 mm (2) 0.75 mm (3) 0.80 mm (4) 0.70 mm

**[JEE-Mains 2016]**

**Sol.**(3)

(1) $92 \pm 3 \mathrm{s}$

(2) $92 \pm 2 \mathrm{s}$

(3) $92 \pm 5.0 \mathrm{s}$

(4) $92 \pm 1.8 \mathrm{s}$

**[JEE-Mains 2016]**

**Sol.**(2)

$\mathrm{T}_{\mathrm{AV}}=92 \mathrm{s}$

$(|\Delta \mathrm{T}|)_{\text {mean }}=1.5 \mathrm{s}$

Least conunt of clock is 1sec. So value that clock can measure should be 1, 2, 3……so on

so reported mean time should be

$92 \pm 2$

Ref : NCERT (XIth) Ex. 2.7, Page. 25

Diameter of capilary, $\mathrm{D}=1.25 \times 10^{-2} \mathrm{m}$

rise of water, $h=1.45 \times 10^{-2} \mathrm{m}$

Using $\mathrm{g}=9.80 \mathrm{m} / \mathrm{s}^{2}$ and the simplified relation

$\mathrm{T}=\frac{\mathrm{rhg}}{2} \times 10^{3} \mathrm{N} / \mathrm{m},$ the possible error in surface tension is closest to :

(1) 2.4% (2) 10% (3) 0.15% (4) 1.5%

**[JEE-Mains 2017]**

**Sol.**(4)

$\mathrm{T}=\frac{\mathrm{rhg}}{2} \times 10^{3}$

$\frac{\Delta \mathrm{T}}{\mathrm{T}}=\frac{\Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta \mathrm{h}}{\mathrm{h}}+0$

$100 \times \frac{\Delta \mathrm{T}}{\mathrm{T}}=\left(\frac{10^{-2} \times .01}{1.25 \times 10^{-2}}+\frac{10^{-2} \times .01}{1.45 \times 10^{-2}}\right) 100$

$=(0.8+0.689)$

$=(1.489)$

$100 \times \frac{\Delta \mathrm{T}}{\mathrm{T}}=1.489 \%$

= 1.5%

(1) 3.5 % (2) 4.5 % (3) 6 % (4) 2.5 %

**[JEE-Mains 2018]**

**Sol.**(2)

Density $=\frac{\text { Mass }}{\text { Volume }}$

$\frac{1 \Delta \mathrm{d}}{\mathrm{d}}=\frac{1 \Delta \mathrm{M}}{\mathrm{M}}+\frac{3 \Delta \mathrm{L}}{\mathrm{L}}$

= 1.5 + 3(1)

= 4.5 %