33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020
Share Embed Flag
Download ePaper

33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

All previous year questions , and these are downloaded from the sources in the internet. I don't own these resources and these are copyrighted by MTG. All previous year questions , and these are downloaded from the sources in the internet. I don't own these resources and these are copyrighted by MTG.

Vicky18
24.06.2022 Views

Telegram @unacademyplusdiscounts78 NEET-AIPMT Chapterwise Topicwise Solutions PhysicsANSWER KEY1. (b) 2. (d) 3. (d) 4. (a) 5. (c) 6. (d) 7. (a) 8. (b) 9. (c) 10. (b)Hints & Explanations1. (b)2. (d) : Given : initial length = L, area of cross section = ANew length after mass M is suspended on the wire = L\ Change in length, DL = L 1 – L.Now Young’s modulus, Y = Stress= mgAL MgL∆Lor AL ( 1 − L)3. (d) : Stress = F Mg=A A,Strain = ∆L L l L l= + − =L L LEnergy stored in the wire is,Strain = FA×L∆LU = 1 × Stress × Strain × Volume2= 1 2 × Mg× lA L × A×L = 1 2 MglFl4. (a) : Young’s modulus, Y =A ∆ lSince initial volume of wires are same and their areasof cross sections are A and 3A so lengths are 3l and lrespectively.For wire 1,F∆l = ⎛ l⎝ ⎜ ⎞AY ⎠⎟ 3...(i)For wire 2, let F ′ force is appliedF′ =A Y ∆ l3 l⎛ F′⎞⇒ ∆l =⎝⎜3 AY ⎠⎟ lFrom eqns (i) and (ii),⎛ F ⎞ ⎛ F′⎞l⎝⎜AY ⎠⎟ 3 =⎝⎜AY ⎠⎟ l ⇒ F ′ = 93F...(ii)5. (c) : Bulk modulus B is given aspVB = − ...(i)∆ VThe volume of a spherical object of radius r is given as4 4V = πr 3 , ∆V = π( 3r 2 ) ∆r3 34 3πrV∴ − =3V ror − =−∆V4 ∆V∆rr ∆r3 3 23πprPut this value in eqn. (i), we get B =− 3∆r∆r pFractional decrease in radius is − =r 3B6. (d) : Let L and A be lengthand area of cross section of eachwire. In order to have the lowerends of the wires to be at thesame level (i.e. same elongationis produced in both wires), letweights W s and W b are added to steel and brass wiresrespectively. Then, by definition of Young’s modulus, theelongation produced in the steel wire isWL⎛s∆Ls= as Y W / A ⎞=YA⎝⎜LL⎠⎟∆ /sWL band that in the brass wire is ∆Lb=YA bBut DL s = DL b (given)WL s WL b WsYs∴ = or =YA s YA b WbYbAs Y sY= 2 ; ∴ Ws= 2b Wb17. (a) : Depth of ocean, d = 2700 mDensity of water, r = 10 3 kg m –3Compressibility of water, K = 45.4 × 10 –11 Pa –1∆VV = ?Excess pressure at the bottom, DP = rgd= 10 3 × 10 × 2700 = 27 × 10 6 Pa∆PWe know,B =( ∆VV/ )⎛ ∆V⎞ ∆PK ∆P K⎝⎜V ⎠⎟ = B= ⎛= 1 ⎞.⎝⎜B ⎠⎟= 45.4 × 10 –11 × 27 × 10 6 = 1.2 × 10 –28. (b) : As V = Al ... (i)where A is the area of cross-section of the wire.( F/ A)FlYoung’s modulus, Y = =( ∆l/ l)A∆l2Fl Fl∆l = YA=(Using (i))YV Dl ∝ l 2Hence, the graph between Dl and l 2 is a straight line.

Mechanical Properties of SolidsTelegram @unacademyplusdiscounts799. (c) : Young’s modulus,FL FLFLY = = 4 or ∆L=4A∆L2 2πD∆LπDYwhere F is the force applied, L is the length, D is thediameter and DL is the extension of the wire respectively.As each wire is made up of same material therefore theirYoung’s modulus is same for each wire.For all the four wires, Y, F (= tension) are the same.∴In (a)∆L ∝In (b)In (c)LD 2L 200 cm 3 −1D 2 =022 = 5×10 cm(. cm)L 300 cm3 −1D 2 =032 = 33 . × 10 cm(. cm)L 50 cm3 −1D 2 =0052 = 20 × 10 cm(. cm)In (d)L 100 cm3 −1D 2 =012 = 10 × 10 cm(. cm)Hence, DL is maximum in (c).FL 4FL10. (b) : As Y = =A∆L2π D ∆LFL∆L = 4 2πDY2∆LSFSLSDCYC∴ =∆LCFCL 2C D YS Swhere subscripts S and C refer to copperand steel respectively.Here, F S = (5m + 2m)g = 7mgF C = 5mgLSqD SpY S= , = , = sLCDCYC∴ = ⎛ 2∆LS7mg⎝ ⎜ ⎞ ⎛ 1⎞⎠⎟⎝⎜⎠⎟ ⎛ ⎝ ⎜ 1q⎞ ⎠ ⎟ 7q() =∆LC5mgp s 25psSteelCoppervvv

Mechanical Properties of Solids

Telegram @unacademyplusdiscounts

79

9. (c) : Young’s modulus,

FL FL

FL

Y = = 4 or ∆L

=

4

A∆L

2 2

πD

∆L

πDY

where F is the force applied, L is the length, D is the

diameter and DL is the extension of the wire respectively.

As each wire is made up of same material therefore their

Young’s modulus is same for each wire.

For all the four wires, Y, F (= tension) are the same.

∴

In (a)

∆L ∝

In (b)

In (c)

L

D 2

L 200 cm 3 −1

D 2 =

02

2 = 5×

10 cm

(. cm)

L 300 cm

3 −1

D 2 =

03

2 = 33 . × 10 cm

(. cm)

L 50 cm

3 −1

D 2 =

005

2 = 20 × 10 cm

(. cm)

In (d)

L 100 cm

3 −1

D 2 =

01

2 = 10 × 10 cm

(. cm)

Hence, DL is maximum in (c).

FL 4FL

10. (b) : As Y = =

A∆L

2

π D ∆L

FL

∆L = 4 2

πDY

2

∆LS

FS

LS

DC

YC

∴ =

∆LC

FC

L 2

C D Y

S S

where subscripts S and C refer to copper

and steel respectively.

Here, F S = (5m + 2m)g = 7mg

F C = 5mg

LS

q

D S

p

Y S

= , = , = s

LC

DC

YC

∴ = ⎛ 2

∆LS

7mg

⎝ ⎜ ⎞ ⎛ 1⎞

⎠

⎟

⎝

⎜

⎠

⎟ ⎛ ⎝ ⎜ 1

q

⎞ ⎠ ⎟ 7q

() =

∆LC

5mg

p s 2

5ps

Steel

Copper

vvv

logo

products

Resources

Company

Terms of service
Privacy policy
Cookie policy
Cookie settings
Imprint
Change language
Made with love in Switzerland
© 2024 Yumpu.com all rights reserved

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!