2020 JUPEB PHYSICS QUESTIONS AND ANSWERS (OBJECTIVE AND THEORY)

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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