Physics NEET Questions — Free MCQ Practice

Physics MCQs for NEET — Practice Questions with Answers

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The units of modulus of rigidity are

$N - m$

$N / m$

$N - m^{2}$

$N / m^{2}$

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Explanation

$N / m^{2}$

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The unit of absolute permittivity is

$F m$ (Farad-metre)

$F m^{- 1}$ (Farad/metre)

$F m^{- 2}$ (Farad/ $m e t r e^{2}$ )

F (Farad)

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Explanation

$C = \frac{ε_{0} A}{d} \Rightarrow ε_{0} = \frac{C d}{A} = F m^{- 1}$ (Farad/meter)

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List-I

I. Joule

II. Watt

III. Volt

IV. Coulomb

List-II

A. Henry × Amp/sec

B. Farad × Volt

C. Coulomb × Volt

D. Oersted × cm

E. Amp × Gauss

F. Amp² × Ohm

Codes: [Only for droppers]

$I - A, I I - F, I I I - E, I V - D$

$I - C, I I - F, I I I - A, I V - B$

$I - C, I I - F, I I I - A, I V - E$

$I - B, I I - F, I I I - A, I V - C$

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Explanation

$I - C, I I - F, I I I - A, I V - B$

Checking for options of List -i

I. $P o t e n t i a l = \frac{W o r k}{C h a r g e} W o r k = P o t e n t i a l X C h a r g e J o u l e = C o u l o m b \times V o l t$

$I I . P o w e r = {(C u r r e n t)}^{2} \times R e s i s \tan c e W a t t = A m p e r e^{2} \times O h m$

$I I I . V o l t a g e (V) = \frac{L i}{t} w h e r e L : I n d u c t o r (U n i t : H e n r y) i : C u r r e n t P u t t i n g u n i t s, V o l t = \frac{H e n r y \times A m p e r e}{S e c o n d}$

IV. $F o r a c a p a c i t o r, Q = C \times V C o u l o m b = F a r a d X V o l t$

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Which relation is wrong

1 Calorie = 4.18 Joules

$1 Å = 10^{- 10} m$

$1 M e V = 1 .6 \times 10^{- 13} J o u l e s$

$1 N e w t o n = 10^{- 5} D y n e s$

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Explanation

1 Newton = $10^{5}$ Dyne

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If $x = a t + b t^{2}$ , where x is the distance travelled by the body in kilometres while t is the time in seconds, then the units of b are

$k m / s$

$k m - s$

$k m / s^{2}$

$k m - s^{2}$

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Explanation

$[x] = [b t^{2}] \Rightarrow [b] = [x / t^{2}] = k m / s^{2}$

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The equation $(P + \frac{a}{V^{2}})$ $(V - b)$ = constant. The units of a are

$D y n e \times c m^{5}$

$D y n e \times c m^{4}$

$D y n e / c m^{3}$

$D y n e / c m^{2}$

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Explanation

Using dimensional analysis, addition is only possible for quantitites having same dimensions

$[P] = [\frac{a}{V^{2}}] ∴ [a] = [P V^{2}] = \frac{F o r c e}{A r e a} \times {(V o l u m e)}^{2} = F o r c e \times \frac{{(L^{3})}^{2}}{L^{2}} = F o r c e X {(L e n g t h)}^{4} = d y n e \times c m^{4}$

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Which of the following quantity is expressed as force per unit area

Work

Pressure

Volume

Area

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Explanation

$P r e s s u r e = \frac{F o r c e}{A r e a}$

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Unit of impulse is

Newton

$k g - m$

$k g - m / s$

Joule

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Explanation

Impulse = Force × time $= (k g - m / s^{2}) \times s = k g - m / s$

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Which is not a unit of electric field

$N C^{- 1}$

$V m^{- 1}$

$J C^{- 1}$

$J C^{- 1} m^{- 1}$

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Explanation

${JC}^{- 1} is the unit of electric potential .$

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The correct value of $0^{o} C$ on the Kelvin scale is

$273 .15 K$

$272 .85 K$

$273 K$

$273 .2 K$

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Explanation

$K = C + 273 .15$

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