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Question 1 Report
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Question 2 Report
The figure above shows a uniform circular object, center R and diameter PS. A circular section 6 and Diameter PR is cut from it. if PQRS is a straight line, where is center of gravity of the figure
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Question 3 Report
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Question 4 Report
In the figure above W1 = 200g, and W2 =450g . Calculate the extension of the spring per unit load? [g = 10ms-2]
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Question 5 Report
Eight alpha decays and six beta decays are necessary before an atom of \( {}^{238}_{92}\mathrm{U} \) achieves stability. The final product in the chain has an atomic number of
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92 = (8 X 2) + (6 X -1) + a
a = 82
Question 6 Report
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Question 7 Report
Three resistors, with substances \(250\Omega\), \(500\Omega\) and \(1k\Omega\) are connected in series. A 6V battery is connected to either end of the combination. Calculate the pot3ential difference between the end of the \(250\Omega\) resistor.
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v1 = (R1)RT VT
= 2501750 x 61
= 0.86V
Question 8 Report
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Question 9 Report
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2 x 17 = 34cm
F = Vλ
= 34034×10−2
= 1000Hz
Question 10 Report
Which of the following pairs is NOT part of the electromagnetic spectrum? i. Radio waves. ii. Beta rays. iii. Gamma rays. iv. Alpha rays
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Question 12 Report
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Question 13 Report
An ammeter of resistance \(1.0\,\Omega\) has a full-scale deflection of 50mA. Determine the resultant full-scale deflection of the meter when a shunt of \(0.0111\,\Omega\) is connected across its terminals.
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RT = 0.1×0.01110.1+0.0111
= 9.99 x 10-3Ω
V = IR ⇒ I
= 5×10−39.9×10−3
= 0.5A ; = 500mA
Question 14 Report
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70×12007×273 = 75×V227+273
V2 = 1200cm3
Question 15 Report
The air column as shown above is set into vibration by the turning fork. In this State of resonance, the waves in the air column will be
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Question 16 Report
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Question 17 Report
The magnification of the image of an object placed in front of a convex mirror is \( \frac{1}{3} \). If the radius of curvature of the mirror is 24cm, what is the distance between its object and its image?
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12 = v+3vv+3v
v = 16cm, u = 48cm
distance between U and V
= 48 - 16 =32cm
Question 18 Report
The figure above shows a conductor PQ carrying a current I in the direction shown. At a particular position near the conductor is a compass needle K. Neglecting the earth's magnetic field, the compass needle will
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Question 19 Report
Question 20 Report
A block of mass m is held in equilibrium against a vertical wall by a horizontal force. If the coefficient of friction between the wall and the block is µ, the minimum value of the horizontal force is
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Question 22 Report
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0θ ....325mm
θ .... 190mm
0−θ100−θ = 325−190875−190
θ = -25oC
= 243k
Question 23 Report
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Question 24 Report
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10×A27+273 = L×A100+273
L = 12.43cm
Question 25 Report
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Question 26 Report
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Question 27 Report
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Question 28 Report
In the above circuit diagrams A is the ammeter and V is the Voltmeter. Which of the circuits is correct for finding the correct value of the resistance R
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Question 29 Report
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Question 30 Report
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40 = 20×10input x 1001
input = 500J
work against friction = input - output
= 500 - 200 = 300J
Question 31 Report
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Question 32 Report
A simple pendulum of mass m moves along an arc pf a circle radius R in a vertical plane as shown above. What is the work done by gravity in a downward swing through the angle q to 0 degree
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Question 33 Report
A beam of radiation is passed between a pair of charged plates as indicated in the fig above. Beam P is undetected While Q is deflected to the left. P and Q respectively should be
I.γ-rays β-rays
II. x-rays β-rays
III. γ-rays x-rays
IV. x-rays x-rays
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Question 34 Report
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Question 35 Report
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= 60 x 45 x 10-2
= 27ms-1
Question 36 Report
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Question 37 Report
Three 2-µF capacitors are arranged as seen in the circuit above . The effective capacitance between E and F is
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To find the effective capacitance between points E and F, we need to analyze the circuit and determine which capacitors are in series and which are in parallel. Capacitors in series add reciprocally, while capacitors in parallel add directly.
Starting from point E, the two capacitors connected to it are in series. Their combined capacitance is:
C1 + C2 = 2µF + 2µF = 4µF
This 4µF equivalent capacitor is in parallel with the third 2µF capacitor, so the total capacitance between E and F is:
Ctotal = C1 + C2 || C3
Ctotal = C1 + C2 + C3 / (C1 + C2) || C3
Ctotal = (C1 + C2) * C3 / (C1 + C2 + C3)
Ctotal = (4µF) * (2µF) / (4µF + 2µF)
Ctotal = 8µF / 6µF
Ctotal = 1.33µF
Therefore, the effective capacitance between E and F is 1.33µF.
Question 38 Report
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Question 39 Report
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Question 40 Report
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Question 41 Report
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Question 42 Report
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Question 43 Report
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Question 44 Report
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Question 45 Report
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