2020 Jupeb Physics Answers

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Jupeb Physic-Obj

01-10: ACCDACAABD

11-20: CBAACBBBBC

21-30: DBCBABADBC

31-40: ABCAADACCC

41-50: CDBAACBBBA

Completed.

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(1ai)

Young’s modulus is defined as the ratio of tensile strength to tensile strain. It has the same unit as Stress, N/m².

(1aii)

Stiffness of an elastic material is the force required to produce a unit extension. It gives a measure of how strong the material is.

i.e stiffness, k = Force/Extension in N/m

(b)

(i) Up to the proportional limit,

Force constant, k = force/extension

=200/0.3×10^–2

=66,666.67N/m

Or =66.67KN/m

(ii)

Workdone = 1/2 Fe

=1/2 × 200 × 0.3 × 10^–²

=0.3 Joules

(1ci)

(View image above)

m1• ms• m2•

0.20kg 0.050kg 0.30kg

<–———————0.15m———————>

F31=G×0.050×0.20/(0.15-0.05)²

=G×0.01/0.1²

=G

F32=G×0.050×0.30/(0.05)²

=6G

Gravitational Force acting on m3=6G-G

=5G

Where gravitational constant, G=6.67×10^-¹¹

Force acting on third particle =5×6.67×10^-¹¹

=3.335×10^-10N

(1cii)

For gravitational force to be zero.

F31-F32=0

G×0.050×0.20/r² = G×0.050×0.30/(0.15-r)²

0.30r²=0.20(0.15–r)²

0.30r² = 0.20(0.0225–0.3r+r²)

0.30r² = 0.0045–0.06r+0.20r²

0.1r²+0.06r–0.0045=0

r²+0.6r–0.045=0

Using formula:

r= –y±√0.36 + 0.18 / 2

=–0.6±√0.54/2

For positive r, r=–0.6 + 0.7348/2

=0.0674

It should be placed 0.0674m from the 0.20kg particle.

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(4ai)

(i) Frequency

(ii) Period

(iii) Wavelength

(iv) Velocity(speed)

(v) Amplitude(maximum displacement)

(4aii)

Linear Magnification is defined as the ratio of of object image to object length, mathematicall expressed as

M=u/v

Where, u= the object image

v= is the object length

(4b) & (4c)

(View image above)

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(6ai)

Flemming’s left hand rule states that when a current carrying conductor is placed in an external magnetic field, the conductor experiences a force perpendicular to both the field and the direction of the current flow.

(6aii)

(i) Magnitude of the charge

(ii) The velocity of the particle

(6b)

£=V/r

£=5.2 × 10³ / 4 × 10–²

£=1.3 × 10^³+²

£=1.3 × 10^5 vm^-¹

(6ci)

Np=400turns

Ns=1200turns

Ep=150V

Es=?

Ns/Np = Es/Ep

Es=Ns×Ep/Np

Es=1200×150/400

Es=450V

(6cii)

Ip=?

Is=5A

Es/Ep=Ip/Is

Ip=Es×Is/Ep

Ip=450×5/150

=15A

(6ciii)

p=IpVp=15×150

p=2250W

(6civ)

p=IsVs

=5×450

=2250W

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Completed.