Physics Syllabus (JEE Main)
Q. If the viscosity of a liquid is doubled, how does it affect the flow rate through a pipe?
A.
Flow rate doubles
B.
Flow rate halves
C.
Flow rate remains the same
D.
Flow rate quadruples
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Solution
According to Poiseuille's law, if viscosity is doubled, the flow rate is halved, assuming all other factors remain constant.
Correct Answer: B — Flow rate halves
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Q. If the viscosity of a liquid is doubled, what happens to the flow rate through a pipe, assuming all other factors remain constant?
A.
Flow rate doubles
B.
Flow rate halves
C.
Flow rate remains the same
D.
Flow rate quadruples
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Solution
According to Poiseuille's law, if viscosity is doubled, the flow rate will be halved, assuming other factors remain constant.
Correct Answer: B — Flow rate halves
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Q. If the voltage across a conductor is 15 volts and the current is 3 amperes, what is the resistance?
A.
5 Ω
B.
10 Ω
C.
15 Ω
D.
20 Ω
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Solution
Using Ohm's Law, R = V / I = 15 V / 3 A = 5 Ω.
Correct Answer: A — 5 Ω
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Q. If the voltage across a resistor is doubled, what happens to the current through the resistor, assuming resistance remains constant?
A.
Doubles
B.
Halves
C.
Remains the same
D.
Increases by a factor of four
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Solution
According to Ohm's Law (V = IR), if voltage is doubled and resistance remains constant, current also doubles.
Correct Answer: A — Doubles
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Q. If the voltage across a resistor is doubled, what happens to the current through it?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
According to Ohm's law, if voltage is doubled and resistance remains constant, current also doubles.
Correct Answer: A — It doubles
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Q. If the voltage across a resistor is tripled, what happens to the current through it, assuming resistance remains constant?
A.
It triples.
B.
It doubles.
C.
It remains the same.
D.
It decreases.
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Solution
According to Ohm's Law (I = V/R), if voltage (V) is tripled and resistance (R) remains constant, the current (I) will also triple.
Correct Answer: A — It triples.
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Q. If the volume of a gas is doubled while keeping the temperature constant, what happens to the pressure?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
According to Boyle's Law, if the volume is doubled, the pressure is halved.
Correct Answer: B — It halves
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Q. If the volume of a gas is halved while keeping the temperature constant, what happens to the pressure?
A.
Pressure is halved
B.
Pressure remains constant
C.
Pressure doubles
D.
Pressure quadruples
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Solution
According to Boyle's Law, if the volume of a gas is halved at constant temperature, the pressure will double.
Correct Answer: C — Pressure doubles
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Q. If the wavelength of a wave is halved, what happens to its frequency?
A.
It remains the same
B.
It doubles
C.
It halves
D.
It quadruples
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Solution
Frequency is inversely proportional to wavelength. If the wavelength is halved, the frequency doubles.
Correct Answer: B — It doubles
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Q. If the wavelength of light in a vacuum is 600 nm, what is its wavelength in glass (n = 1.5)?
A.
400 nm
B.
600 nm
C.
900 nm
D.
300 nm
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Solution
The wavelength in a medium is given by λ' = λ/n. Thus, λ' = 600 nm / 1.5 = 400 nm.
Correct Answer: A — 400 nm
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Q. If the wavelength of light in air is 600 nm, what is its wavelength in glass (n=1.5)?
A.
400 nm
B.
450 nm
C.
600 nm
D.
900 nm
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Solution
Wavelength in glass (λ') = λ/n = 600 nm / 1.5 = 400 nm.
Correct Answer: A — 400 nm
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Q. If the wavelength of light is halved, what happens to the frequency of the light?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
Frequency (f) is inversely proportional to wavelength (λ). If λ is halved, f doubles.
Correct Answer: A — It doubles
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Q. If the wavelength of light used in a diffraction experiment is halved, what happens to the position of the first diffraction minimum?
A.
It moves closer to the center
B.
It moves further from the center
C.
It remains unchanged
D.
It disappears
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Solution
Halving the wavelength results in the first minimum moving closer to the center, as the position of minima is directly related to the wavelength.
Correct Answer: A — It moves closer to the center
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Q. If the wavelength of light used in a diffraction experiment is halved, what happens to the position of the minima?
A.
They move closer together
B.
They move further apart
C.
They remain unchanged
D.
They disappear
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Solution
Halving the wavelength causes the minima to move closer together, as the angle for minima is directly proportional to the wavelength.
Correct Answer: A — They move closer together
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Q. If the wavelength of light used in a diffraction experiment is halved, what happens to the angular position of the first minimum in a single-slit diffraction pattern?
A.
It remains the same
B.
It doubles
C.
It halves
D.
It quadruples
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Solution
Halving the wavelength will halve the angle for the first minimum, as the position of minima is directly proportional to the wavelength.
Correct Answer: C — It halves
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Q. If the wavelength of light used in a diffraction experiment is increased, what happens to the diffraction pattern?
A.
It becomes sharper
B.
It becomes broader
C.
It remains unchanged
D.
It disappears
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Solution
Increasing the wavelength results in a broader diffraction pattern as the angles for minima and maxima increase.
Correct Answer: B — It becomes broader
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Q. If the wavelength of light used in a double-slit experiment is 600 nm and the distance between the slits is 0.3 mm, what is the distance between the first and second bright fringes on the screen placed 2 m away?
A.
0.4 m
B.
0.6 m
C.
0.8 m
D.
0.2 m
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Solution
Distance between fringes = (λD)/d = (600 x 10^-9 m * 2 m) / (0.3 x 10^-3 m) = 0.004 m = 0.4 m.
Correct Answer: A — 0.4 m
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Q. If the wavelength of light used in a double-slit experiment is increased, what happens to the position of the interference fringes?
A.
Fringes move closer together
B.
Fringes move further apart
C.
Fringes disappear
D.
Fringes become brighter
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Solution
Increasing the wavelength increases the fringe width, causing the fringes to move further apart.
Correct Answer: B — Fringes move further apart
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Q. If the wavelength of light used in an interference experiment is 500 nm, what is the fringe separation when the screen is placed 2 m away from the slits separated by 0.1 mm?
A.
0.01 m
B.
0.025 m
C.
0.05 m
D.
0.1 m
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Solution
Fringe separation β = λD/d = (500 x 10^-9 m)(2 m)/(0.1 x 10^-3 m) = 0.01 m.
Correct Answer: C — 0.05 m
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Q. If the wavelength of light used in Young's experiment is 600 nm and the distance between the slits is 0.1 mm, what is the distance between the first and second bright fringes on a screen 2 m away?
A.
0.12 m
B.
0.24 m
C.
0.36 m
D.
0.48 m
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Solution
Fringe separation (β) = λD/d. β = (600 x 10^-9 * 2) / 0.0001 = 0.012 m. Distance between first and second bright fringes = 2β = 0.024 m.
Correct Answer: B — 0.24 m
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Q. If the wavelength of light used in Young's experiment is 600 nm, what is the fringe width when the distance between the slits is 0.1 mm and the distance to the screen is 2 m?
A.
0.03 mm
B.
0.06 mm
C.
0.12 mm
D.
0.15 mm
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Solution
Fringe width (β) = (λD)/d = (600 x 10^-9 * 2)/(0.1 x 10^-3) = 0.012 mm = 0.06 mm.
Correct Answer: B — 0.06 mm
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Q. If the Wheatstone bridge is balanced, what is the potential difference across the galvanometer?
A.
Maximum.
B.
Minimum.
C.
Zero.
D.
Equal to the supply voltage.
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Solution
When the bridge is balanced, the potential difference across the galvanometer is zero.
Correct Answer: C — Zero.
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Q. If the Wheatstone bridge is unbalanced, what can be inferred about the potential difference across the galvanometer?
A.
It is zero
B.
It is positive
C.
It is negative
D.
It is non-zero
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Solution
An unbalanced Wheatstone bridge will have a non-zero potential difference across the galvanometer.
Correct Answer: D — It is non-zero
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Q. If the Wheatstone bridge is unbalanced, what can be inferred about the resistances?
A.
R1/R2 = R3/R4
B.
R1/R2 ≠ R3/R4
C.
R1 + R2 = R3 + R4
D.
R1 - R2 = R3 - R4
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Solution
An unbalanced Wheatstone bridge indicates that the ratio of the resistances is not equal.
Correct Answer: B — R1/R2 ≠ R3/R4
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Q. If the Wheatstone bridge is unbalanced, what can be said about the potential difference across the galvanometer?
A.
It is zero.
B.
It is maximum.
C.
It is equal to the supply voltage.
D.
It is constant.
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Solution
In an unbalanced bridge, there is a potential difference across the galvanometer, leading to maximum current flow.
Correct Answer: B — It is maximum.
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Q. If the Wheatstone bridge is unbalanced, what happens to the current through the galvanometer?
A.
It becomes zero.
B.
It increases.
C.
It decreases.
D.
It becomes infinite.
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Solution
In an unbalanced bridge, there is a potential difference across the galvanometer, causing current to flow.
Correct Answer: B — It increases.
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Q. If the work function of a material is 2.5 eV, what is the minimum wavelength of light required to emit photoelectrons?
A.
400 nm
B.
500 nm
C.
600 nm
D.
700 nm
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Solution
Minimum wavelength (λ) = hc / Φ = (1240 nm·eV) / 2.5 eV = 496 nm.
Correct Answer: B — 500 nm
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Q. If the work function of a metal is 2.5 eV, what is the minimum wavelength of light required to emit photoelectrons?
A.
400 nm
B.
500 nm
C.
600 nm
D.
700 nm
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Solution
Minimum wavelength (λ) = hc / Φ = (1240 nm·eV) / 2.5 eV = 496 nm.
Correct Answer: B — 500 nm
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Q. If the work function of a metal is 4.5 eV, what is the threshold wavelength for the photoelectric effect?
A.
400 nm
B.
500 nm
C.
600 nm
D.
700 nm
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Solution
The threshold wavelength can be calculated using the equation λ = hc/φ. Substituting h = 4.14 x 10^-15 eV·s, c = 3 x 10^8 m/s, and φ = 4.5 eV gives λ ≈ 400 nm.
Correct Answer: A — 400 nm
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Q. If the Young's modulus of a material is 100 GPa and it is subjected to a tensile stress of 200 MPa, what is the strain produced?
A.
0.002
B.
0.0025
C.
0.01
D.
0.005
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Solution
Strain = Stress / Young's modulus = 200 MPa / 100 GPa = 0.002.
Correct Answer: A — 0.002
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