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NEET NEET Questions — Free MCQ Practice

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A gas undergoes a process in which its pressure P and volume V are related as ${VP}^{n} = constant$ . The bulk modulus for the gas in the process is
[Only for Dropper and XII batch]

A.

$\frac{P}{n}$

B.

$P^{1 / n}$

C.

nP

D.

$P^{n}$

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Explanation

$(3) D i f f e r e n t i a t e V P^{n} = C w . r . t o V u \sin g m u l t i p l i c a t i o n r u l e : V \frac{d}{d V} (P^{n}) + P^{n} \frac{d}{d V} (V) = 0 V n P^{n - 1} \frac{d P}{d V} + P^{n} . 1 = 0 V n P^{n - 1} d P + P^{n} . d V = 0 n V P^{n - 1} d P = - P^{n} . d V \Rightarrow \frac{d P}{d V} = - \frac{P^{n}}{n V P^{n - 1}} = - \frac{P}{n V} B u l k M o d u l u s = - \frac{d P}{\frac{d V}{V}} = - (- \frac{P}{\frac{n V}{V}}) = \frac{P}{n}$

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The breaking stress of a wire depends upon

A.

material of the wire

B.

length of the wire

C.

radius of the wire

D.

shape of the cross section

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Explanation

Breaking stress depends only on nature of material

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The elastic energy stored in a wire of Young's Modulus Y is -

A.

$Y \times \frac{{strain}^{2}}{volume}$

B.

$stress \times strain \times volume$

C.

$\frac{{strain}^{2} \times volume}{2 Y}$

D.

$\frac{1}{2} \times s t r e s s \times s t r a i n \times v o l u m e$

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Explanation

$U = \frac{1}{2} F ∆ l = \frac{1}{2} σ A ∆ l = \frac{1}{2} σ \frac{∆ l}{l} A l = \frac{1}{2} σ ε V = \frac{1}{2} \times s t r e s s \times s t r a i n \times v o l u m e$

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If Young modulus (Y) equal to bulk modulus (B). Then the Poisson ratio is :

A.

$\frac{1}{3}$

B.

$\frac{2}{3}$

C.

$\frac{1}{2}$

D.

$\frac{1}{4}$

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Explanation

$Y = 3 B (1 - 2 σ)$

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The bulk modulus of a spherical object is B. If it is subjected to uniform pressure p, the fractional decrease in radius is

A.

$\frac{P}{B}$

B.

$\frac{B}{3 p}$

C.

$\frac{3 p}{B}$

D.

$\frac{p}{3 B}$

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Explanation

(d)The object is spherical and the bulk modulus is represented by B. It is the ratio of normal stress to the volumetric strain.

Hence $B = \frac{F / A}{∆ V / V}$

$\Rightarrow \frac{∆ V}{V} = \frac{P}{B} \Rightarrow |\frac{∆ V}{V}| = \frac{P}{B}$

Hence p is applied pressure on the object and $\frac{∆ V}{V}$ is

volume strain

Fractional decrease in volume

$\Rightarrow \frac{∆ V}{V} = 3 \frac{∆ R}{R} [S o, V = \frac{4}{3} {πR}^{3}]$

Volume of the sphere decreases due to the decrease in its radius.

Hence $\frac{∆ V}{V} = \frac{3 ∆ R}{R} = \frac{P}{B} \Rightarrow \frac{∆ R}{R} = \frac{P}{3 B}$

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The Young's modulus of steel is twice that of brass. Two wires of same length and of same area of cross-section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weight added to the steel and brass wires must be in the ratio of

A.

1:2

B.

2:1

C.

4:1

D.

1:1

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Explanation

Given, Y_steel=2Y_brass and L_s=L_b and A_s=A_b

such that ΔL_s=ΔL_b

As we know,Young's modulus

Y=stress/strain=W/A/ΔL/L

i.e. W_s/W_b=Y_s/Y_b=2Y_b/Y_b=2/1

=> 2:1

Thus, weight added to the steel and brass wires must be in the ratio of 2:1

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Copper of fixed volume V is drawn into wire of length l. When this wire is subjected to a constant force F, the extension produced in the wire is Δl. Which of the following graphs is a straight line?

A.

Δl versues 1/l

B.

Δl versus l²

C.

Δl versus 1/l²

D.

Δl versus l

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Explanation

For Hooke's law, Young Modulus

y=F/A/Δl/l

=F x l/Δl x A ...(i)

V = A x l=constant ...(ii)

From Eqs. (i) and (ii),

y=F x lxl / Δl x V

Δl=F/V x y x l²

=>Δl∝l²

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The following four wires are made of the same material. Which of then will have the largest extension when the same tension is applied?

A.

Length=50cm, diameter=0.5mm

B.

Length=100cm, diameter=1mm

C.

Length=200cm, diameter=2mm

D.

Length=300cm, diameter=3mm

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Explanation

(a) ΔL=FL/AY or ΔL∝L/d² [∴ A=πd²/4]

Therefore,ΔL will be maximum for that wire for which L/A is maximum.

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The Young's modulus of a wire of length L and radius r is $Y N / m^{2} .$ If the length and radius are reduced to L/2 and r/2, then its Young's modulus will be -

A.

1:1

B.

1:4

C.

1:8

D.

8:1

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Explanation

(a) Young's modulus of wire does not varies with dimension of wire. It is the property of given material.

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When a certain weight is suspended from a long uniform wire, its length increases by one cm. If the same weight is suspended from another wire of the same material and length but having a diameter half of the first one then the increase in length will be -

A.

0.5 cm

B.

2 cm

C.

4 cm

D.

8 cm

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Explanation

(c)

$l = \frac{F L}{A Y} \Rightarrow l \propto \frac{1}{r^{2}} (F, L a n d Y a r e c o n s \tan t) \frac{l_{2}}{l_{1}} = {(\frac{r_{1}}{r_{2}})}^{2} = {(2)}^{2} = 4 \Rightarrow l_{2} = 4 l_{1} = 4 c m$

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