datavorous/entrance-exam-dataset huggingface dataset:
An ac source is connected in given series LCR circuit. The rms potential difference across the capacitor of $20 \mu \mathrm{F}$ is _____V.

Options:

The correct answer is:
50

Solution:
$\begin{aligned} & \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=100 \times 1=100 \Omega \\ & \mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{100 \times 20 \times 10^{-6}}=500 \Omega \\ & \mathrm{Z}=\sqrt{\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2+\mathrm{R}^2} \\ & \sqrt{(100-500)^2+300^2} \\ & \mathrm{Z}=500 \Omega \\ & \mathrm{i}_{\text {rms }}=\frac{\mathrm{V}_{\text {rms }}}{\mathrm{Z}}=\frac{50}{500}=0.1 \mathrm{~A} \\ & \text { rms voltage across capacitor } \\ & \mathrm{V}_{\text {rms }}=\mathrm{X}_{\mathrm{C}} \mathrm{i}_{\text {rms }} \\ & =500 \times 0.1=50 \mathrm{~V}\end{aligned}$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2024 (05 Apr Shift 1)
An ac source is connected to a capacitor C. Due to decrease in its operating frequency:

Options:
A capacitive reactance remains constant
B capacitive reactance decreases
C displacement current increases
D displacement current decreases

The correct answer is:
displacement current decreases

Solution:
Capacitive reactance is given by,

XC=1ωC⇒XC∝1ω

Clearly, as ω decreases the value of the XC will increase.

Now, as impedance of the circuit is decreasing and as we know, conduction current will be equal to the displacement current, hence displacement current will decrease.

Topic: Physics, Alternating Current, NEET, NEET 2023 (All India)
An ac source is connected to a resistive circuits. Which of the following is true

Options:
A Current leads the voltage and both are in same phase
B Current lags behind the voltage and both are in same phase
C Current and voltage are in same phase
D Any of the above may be true depending upon the value of resistance

The correct answer is:
Current and voltage are in same phase

Solution:
In a pure ohmic AC Resistance, the current and voltage are both said to be "inphase" as there is no phase difference between them. The current flowing through the resistor is directly proportional to the voltage across it.

Topic: Physics, Alternating Current, JEE Main
An AC source is connected to an inductance of 100 mH, a capacitance of 100 μF and a resistance of 120 Ω as shown in figure. The time in which the resistance having a thermal capacity 2 J C-1∘ will get heated by 16°C is _____ s.

Options:

The correct answer is:
15

Solution:
From the diagram given,

Angular frequency ω=100 rad s-1 and Vrms=V02=2022=20 V.

Now, XL=ωL=100×100×10-3=10 Ω and XC=1ωC=1100×100×10-6=100 Ω

XL-XC=10-100=90 Ω

The impedance of the circuit,

Z=XL-XC2+R2=902+1202=150 Ω

The current in the circuit,

imrs=VrmsZ=20150 A

Power factor cosϕ=RZ

Now

VrmsirmscosϕΔt=msΔT⇒Δt=msΔTVrms×VrmsZ×RZ=msΔTZ2Vrms2×R=2×16×22500400×120=15 s

⇒Δt=15 s

Topic: Physics, Alternating Current, JEE Main, JEE Main 2022 (28 Jun Shift 1)
An ac source of 50 V (r.m.s value) is connected across a series R-C circuit. If the r.m.s voltage across the resistor is 40 V, then the r.m.s voltage across the capacitor is

Options:
A 10 V
B 20 V
C 30 V
D 40 V

The correct answer is:
30 V

Solution:
For an R - C circuit
Applied voltage, V=VR2+VC2
â´ 50=402+VC2
â VC=30 V

Topic: Physics, Alternating Current, JEE Main
An AC source of angular frequency $\omega$ is fed across a resistor $R$ and a capacitor $C$ in series. The current flowing in the circuit found to be $I$. Now the frequency of the source is changed to $\frac{\omega}{3}$, (maintaining the same voltage) the current in the circuit is found to be halved. What is the ratio of reactance to resistance at the original frequency?

Options:
A $\sqrt{\frac{5}{7}}$
B $\sqrt{\frac{3}{4}}$
C $\sqrt{\frac{3}{5}}$
D $\sqrt{\frac{7}{5}}$

The correct answer is:
$\sqrt{\frac{3}{5}}$

Solution:
At angular frequency $\omega$, current through resistance R and capacitance C is given by
$I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{\sqrt{R^2+X_C^2}}=\frac{V_{\mathrm{rms}}}{\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}}$...(i)
When angular frequency is changed to $\frac{\omega}{3}$, then the current becomes
$$
\frac{I_{\mathrm{rms}}}{2}=\frac{V_{\mathrm{rms}}}{\sqrt{R^2+\left(\frac{1}{\frac{\omega}{3} C}\right)^2}}=\frac{V_{\mathrm{rms}}}{\sqrt{R^2+\left(\frac{3}{\omega C}\right)^2}}...(ii)
$$
Dividing Eq. (i) by Eq. (ii), we get
$\begin{array}{rlrl} & & 2=\frac{\sqrt{R^2+\left(\frac{3}{\omega C}\right)^2}}{\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}} \\ & \Rightarrow & 4\left[R^2+\left(\frac{1}{\omega C}\right)^2\right] & =R^2+\left(\frac{3}{\omega C}\right)^2 \\ & \Rightarrow & 3 R^2 & =\frac{5}{\omega^2 C^2} \\ & \Rightarrow & \frac{1}{\omega_C} & =\sqrt{\frac{3}{5}} \\ \Rightarrow & \frac{X_C}{R} & =\sqrt{\frac{3}{5}}\end{array}$

Topic: Physics, Alternating Current, JEE Main
An AC source of angular frequency $\omega$ is fed across a resistor $R$ and a capacitor $C$ in series. The current flowing in the circuit found to be $I$. Now the frequency of the source is changed to $\frac{\omega}{3}$, (maintaining the same voltage) the current in the circuit is found to be halved. What is the ratio of reactance to resistance at the original frequency?

Options:
A $\sqrt{\frac{5}{7}}$
B $\sqrt{\frac{3}{4}}$
C $\sqrt{\frac{3}{5}}$
D $\sqrt{\frac{7}{5}}$

The correct answer is:
$\sqrt{\frac{3}{5}}$

Solution:
At angular frequency $\omega$, current through resistance R and capacitance C is given by
$I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{\sqrt{R^2+X_C^2}}=\frac{V_{\mathrm{rms}}}{\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}}$...(i)
When angular frequency is changed to $\frac{\omega}{3}$, then the current becomes
$$
\frac{I_{\mathrm{rms}}}{2}=\frac{V_{\mathrm{rms}}}{\sqrt{R^2+\left(\frac{1}{\frac{\omega}{3} C}\right)^2}}=\frac{V_{\mathrm{rms}}}{\sqrt{R^2+\left(\frac{3}{\omega C}\right)^2}}...(ii)
$$
Dividing Eq. (i) by Eq. (ii), we get
$\begin{array}{rlrl} & & 2=\frac{\sqrt{R^2+\left(\frac{3}{\omega C}\right)^2}}{\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}} \\ & \Rightarrow & 4\left[R^2+\left(\frac{1}{\omega C}\right)^2\right] & =R^2+\left(\frac{3}{\omega C}\right)^2 \\ & \Rightarrow & 3 R^2 & =\frac{5}{\omega^2 C^2} \\ & \Rightarrow & \frac{1}{\omega_C} & =\sqrt{\frac{3}{5}} \\ \Rightarrow & \frac{X_C}{R} & =\sqrt{\frac{3}{5}}\end{array}$

Topic: Physics, Alternating Current, AP EAMCET, AP EAMCET 2021 (25 Aug Shift 2)
An a.c. source of angular frequency ' $\omega$ ' is fed across a resistor ' $\mathrm{R}$ ' and a capacitor ' $\mathrm{C}$ ' in series. The current registered is I. If now the frequency of source is changed to $\frac{\omega}{3}$ (but maintaining the same voltage), the current in the circuit is found to be halved. The ratio of reactance to resistance at the original frequency ' $\omega$ ' will be

Options:
A $\sqrt{\frac{2}{5}}$
B $\sqrt{\frac{1}{5}}$
C $\sqrt{\frac{4}{5}}$
D $\sqrt{\frac{3}{5}}$

The correct answer is:
$\sqrt{\frac{3}{5}}$

Solution:
$\begin{aligned} & \text { Initial current } I=\frac{V}{Z}, \text { Final current } I^{\prime}=\frac{V}{Z^{\prime}} \\ & \frac{1}{2}=\frac{I^{\prime}}{I}=\frac{Z}{Z^{\prime}} \quad \therefore \frac{Z}{Z^{\prime}}=\frac{1}{2} \\ & \therefore \frac{Z^2}{Z^{\prime 2}}=\frac{1}{4} \quad \therefore \frac{R^2+X_c^2}{R^2+X_c^{\prime 2}}=\frac{1}{4} \\ & \therefore 4 R^2+4 X_c^2=R^2+X_c^{\prime 2} \\ & \therefore 3 R^2=X_c^{\prime 2}-4 X_c^2 \quad \ldots .(1) \\ & X_c=\frac{1}{\omega C}, X^{\prime}=\frac{3}{\omega C} \quad \therefore X^{\prime}=3 X_c \\ & \text { Putting this value of } X^{\prime} \text { in }(1) \text { we get } \\ & 3 R^2=9 X_c^2-4 X_c^2=5 X_c^2 \\ & \therefore \frac{X_c^2}{R^2}=\frac{3}{5}=0.6 \\ & \therefore \frac{X_c}{R}=\sqrt{\frac{3}{5}}\end{aligned}$

Topic: Physics, Alternating Current, MHT CET, MHT CET 2021 (22 Sep Shift 1)
An ac source of angular frequency Ï is fed across a resistor r and a capacitor C in series. The current registered is I. If now the frequency of the source is changed to Ï3 (but maintaining the same voltage), then current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency Ï.

Options:
A 35
B 25
C 15
D 45

The correct answer is:
35

Solution:
At angular frequency ω , the current in RC circuit is given by
irms=VrmsR2+1ωC2......(i)
Also irms2=VrmsR2+1ω3C2=VrmsR2+9ω2C2......(ii)
From equation (i) and (ii) we get
3R2=5ω2C2⇒1ωCR=35 ⇒XCR=35

Topic: Physics, Alternating Current, NEET
An AC voltage is applied to a resistor of resistance \( R \) and an inductor of inductance \( L \) in series. If \( R \) and the inductive reactance are both equal to \( 3 \Omega \), the phase difference between the applied voltage and the current in the circuit is

Options:
A \( \pi / 6 \)
B \( \pi / 4 \)
C \( \pi / 2 \)
D Zero

The correct answer is:
\( \pi / 4 \)

Solution:
Here,
Resistance, R=3 Ω
Inductive reactance, XL=3 Ω
The phase difference between the applied voltage and the current in the circuit is
tanϕ=XLR=3Ω3Ω=1
ϕ = tan - 1 1 or ϕ = π 4

Topic: Physics, Alternating Current, JEE Main
An ac voltage of $10 \sin \omega t$ volt is applied to a pure inducfor of inductance $10 \mathrm{H}$. The current through the inductor in ampere is

Options:
A $\frac{1}{\theta} \sin \left(\omega t-\frac{\pi}{2}\right)$
B $\omega \sin \left(\omega t-\frac{\pi}{2}\right)$
C $\frac{1}{\omega^2} \sin \left(\omega t-\frac{\pi}{2}\right)$
D $\omega^2 \sin \left(\omega t-\frac{\pi}{2}\right)$

The correct answer is:
$\frac{1}{\theta} \sin \left(\omega t-\frac{\pi}{2}\right)$

Solution:
Apply the kirchhoff's loop rule
$$
\begin{aligned}
& V+E=0 \\
& V-L \frac{d i}{d t}=0 \\
& V=L \frac{d i}{d t}
\end{aligned}
$$
$$
\begin{aligned}
& \frac{\mathrm{di}}{\mathrm{dt}}=\frac{\mathrm{V}}{\mathrm{L}}=\mathrm{V}_0 \frac{\sin \omega \mathrm{t}}{\mathrm{L}} \\
& \mathrm{di}=\frac{\mathrm{V}_0}{\mathrm{~L}} \sin \omega \mathrm{t}
\end{aligned}
$$
Integration with respect to time $t$
$$
\begin{aligned}
& i=\frac{V_0}{L}\left(-\frac{\cos \omega t}{\omega}\right)=\frac{-V_0}{\omega L} \cos \omega t \\
& i=\frac{V_0}{\omega L} \sin \left(\omega t-\frac{\pi}{2}\right)=\frac{10}{\omega \times 10} \sin \left(\omega t-\frac{\pi}{2}\right) \\
& i=\frac{1}{\omega} \sin \left(\omega t-\frac{\pi}{2}\right)
\end{aligned}
$$

Topic: Physics, Alternating Current, JEE Main
An A.C. voltage V=5cosâ¡1000t V is applied to an L-R series circuit of inductance 3 mH and resistance 4 Î©. The value of the maximum current in the circuit is ______ A.

Options:

The correct answer is:
1

Solution:
V=5cosâ¡1000t V
V=V0cosâ¡Ït
V0=5 V
Ï=1000 rad s-1
L=3Ã10-3 H, R= 4 Î©
Maximum current I0=V0Z
I0=5Ï2L2+R2=55=1 A

Topic: Physics, Alternating Current, JEE Main
An AC voltage V=20sin200πt is applied to a series LCR circuit which drives a current I=10sin200πt+π3. The average power dissipated is:

Options:
A 21.6 W
B 200 W
C 173.2 W
D 50 W

The correct answer is:
50 W

Solution:
The average power dissipated is,<P>=IVcosϕ

=102×202×cos60°

=50 W

Topic: Physics, Alternating Current, JEE Main, JEE Main 2024 (31 Jan Shift 2)
An ac voltmeter connected between points $\mathrm{A}$ and $\mathrm{B}$ in the circuit below reads $36 \mathrm{~V}$. If it is connected between $\mathrm{A}$ and $\mathrm{C}$, the reading is $39 \mathrm{~V}$. The reading when it is connected between $\mathrm{B}$ and $\mathrm{D}$ is $25 \mathrm{~V}$. What will the voltmeter read when it is connected between A and D?) Assume that the voltmeter reads true rms voltage values and that the source generated a pure ac.)

Options:
A $\sqrt{481} \mathrm{~V}$
B 31V
C 61V
D $\sqrt{3361} \mathrm{~V}$

The correct answer is:
$\sqrt{481} \mathrm{~V}$

Solution:

Topic: Physics, Alternating Current, JEE Main
An accelerated electron has a speed of 5×106 ms-1 with an uncertainty of 0.02%. The uncertainty in finding its location while in motion is x×10-9 m.
The value of x is __________. (Nearest integer)

[Use mass of electron=9.1×10-31 kg,

h=6.63×10-34Js,π=3.14

Options:

The correct answer is:
58

Solution:
Δv=0.02100×5×106=103 m s-1

Δx·Δv=h4πm

x×10-9×103=6.63×10-344×3.14×9.1×10-31

x×10-9×103=0.058×10-3

x=0.058×10-610-9=58

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2021 (25 Jul Shift 2)
An acid base indicator has Ka=3×10-5 . The acid form of the indicator is red and the basic form is blue. By how much must the pH change in order to change the indicator from 75% red to 75% blue (log 3 = 0.4770)

Options:
A 0.95
B 2.3
C 0.75
D 5

The correct answer is:
0.95

Solution:
pH change = pH of 75 % blue - pH of 75% red
=pKa+loglog7525-pKa+loglog2575
=log3+log3
pH=0.954

Topic: Chemistry, Ionic Equilibrium, JEE Main
An acid solution of pH = 6 is diluted 1000 times, the pH of the final solution becomes

Options:
A 6.01
B 9
C 3.5
D 6.99

The correct answer is:
6.99

Solution:
âµ pH = 6
â´H+=10-6 M
After dilution H+=10-61000=10-9 M
â´[H+] from H2O cannot be neglected.
Total H+=10-9+10-7 (neglect common ion effect on H2O)
=10-710-2+1
=10-7(1.01)
pH= -logâ¡1.01Ã10-7
=7-0.0043
=6.9957

Topic: Chemistry, Ionic Equilibrium, JEE Main
An acid solution of $\mathrm{pH}=6$ is diluted 1000 times, the pH of the final solution becomes

Options:
A $6.01$
B 9
C $3.5$
D $6.99$

The correct answer is:
$6.99$

Solution:
$\because \mathrm{pH}=6$
$\therefore$
$\left[\mathrm{H}^{+}\right]=10^{-6} \mathrm{M}$
After dilution $\left[\mathrm{H}^{+}\right]=\frac{10^{-6}}{1000}=10^{-9} \mathrm{M}$
$\therefore\left[\mathrm{H}^{+}\right]$ from $\mathrm{H}_{2} \mathrm{O}$ cannot be neglected.
Total $\left[\mathrm{H}^{+}\right]=10^{-9}+10^{-7}$
$\begin{aligned} &=10^{-7}\left(10^{-2}+1\right) \\ &=10^{-7}(1.01) \\ \mathrm{pH} &=-\log \left(1.01 \times 10^{-7}\right) \\ &=7-0.0043 \\ &=6.9957 \end{aligned}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
An acid type indicator, Hln differes in colour from its conjugate base (In-) . The human eye is sensitive to colour differences only when the ratio [In-]/[HIn] is greater than 10 or smaller than 0.1. What should be the minimum change in the pH of the solution to observe a complete colour change (Ka=1.0×10-5) ?

Options:
A 4
B 2
C 6
D 1

The correct answer is:
2

Solution:
The two conditions when colour of indicator will be visible are derived by
pH=pKa+log[In-][HIn]
(i) pH = 5 + log 10 = 6
(ii) pH = 5 + log 0.1 = 4
Thus, minimum change in pH = 2.

Topic: Chemistry, Ionic Equilibrium, JEE Main
An acidic buffer is obtained on mixing:

Options:
A 100mL of 0.1MCH3COOH and 100mL of 0.1MNaOH
B 100mL of 0.1MHCl and 200mL of 0.1MNaCl
C 100mL of 0.1MCH3COOH and 200mL of 0.1MNaOH
D 100mL of 0.1MHCl and 200mL of 0.1MCH3COONa

The correct answer is:
100mL of 0.1MHCl and 200mL of 0.1MCH3COONa

Solution:
Mixture of weak acid and its salt with strong base acts as buffer solution.

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2020 (03 Sep Shift 1)
An acidic buffer solution can be prepared by mixing solution of

Options:
A Ammonium acetate and acetic acid
B Ammonium chloride and hydrochloric acid
C Sulphuric acid and sodium sulphate
D Acetic acid and sulphuric acid

The correct answer is:
Ammonium acetate and acetic acid

Solution:
An acid buffer solution consists of solution of weak acid with strong base of its salt.

Topic: Chemistry, Ionic Equilibrium, JEE Main
An acidic solution of dichromate is electrolyzed for 8 minutes using 2 A current. As per the following equation

Cr2O72+14H++6e-→2Cr3++7H2O
The amount of Cr3+. obtained was 0.104g. The efficiency of the process (in%) is (Take : F=960000C, At. mass of chromium =52)

Options:

The correct answer is:
60

Solution:
Charge q=it=2×8×60=960 C

96096000=0.01F

Cr2O72-+14H++6e- → 2Cr3+ + 7H2O

0.01F 13×0.01 mole

Theoritical mass of Cr3+=13×60096000×52=0.173g

So, efficiency =WactualWTheoritial×100=0.1040.173×100=60%

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2020 (03 Sep Shift 2)
An active nucleus decays to one-third 13 in 20 Hrs. The fraction of original activity remaining after 80 Hrs is

Options:
A 116
B 181
C 136
D 154

The correct answer is:
181

Solution:
We know, NN0=23t/T

From the given data,

N03N0=1220T

13=1220T ...1

NN0=1280T ...2

From 1 and 2,

∴ NN0=133=181

Topic: Physics, Nuclear Physics, JEE Main
An active nucleus decays to one-third 13 in 20 Hrs. The fraction of original activity remaining after 80 Hrs is

Options:
A 116
B 181
C 136
D 154

The correct answer is:
181

Solution:
We know, NN0=23t/T

From the given data,

N03N0=1220T

13=1220T ...1

NN0=1280T ...2

From 1 and 2,

∴ NN0=133=181

Topic: Physics, Nuclear Physics, TS EAMCET, TS EAMCET 2018 (04 May Shift 1)
An adiabatic vessel contains \( \mathrm{n}_{1}=3 \) mole of diatomic gas. Moment of inertia of each molecule is \( \mathrm{I}=2.76 \mathrm{x} 10^{-46} \mathrm{~kg} \mathrm{~m}^{2} \) and root-mean-square angular velocity is \( \omega_{0}=5 \times 10^{12} \mathrm{rad} / \mathrm{s} \). Another adiabatic vessel contains \( \mathrm{n}_{2}=5 \mathrm{~mole} \) of a monatomic gas at a temperature \( 470 \mathrm{~K} \). Assume gases to be ideal, calculate root-mean-square angular velocity of diatomic molecules when the two vessel are connected by a thin tube of negligible volume. Boltzmann constant \( \mathrm{k}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{molecule} \).

Options:
A \( 6 \times 10^{12} \mathrm{rad} / \mathrm{s} \)
B \( 7 \times 10^{10} \mathrm{rad} / \mathrm{s} \)
C \( 8 \times 10^{10} \mathrm{rad} / \mathrm{s} \)
D \( 10 \times 10^{10} \mathrm{rad} / \mathrm{s} \)

The correct answer is:
\( 6 \times 10^{12} \mathrm{rad} / \mathrm{s} \)

Solution:
We know according to law of equipartition of energy, each gas molecule has 1/2 kT energy associated with each of its degrees of freedom. As a diatomic gas molecule has two rotational degrees of freedom, final temperature of the system is
T fâ¡ = fâ¡ 1 n 1 T 1 + fâ¡ 2 n 2 T 2 fâ¡ 1 n 1 + fâ¡ 2 n 2 ...(i)
Here for diatomic gas,
fâ¡ 1 = 5 , n = 3 and T1 = 250 K
For monatomic gas,
fâ¡ 2 = 3 , n2 = 5 and T2 = 470 K
Thus from Eq. (i),
T fâ¡ = 5 × 3 × 2 5 0 + 3 × 5 × 4 7 0 5 × 3 + 3 × 5 = 3 6 0 K
Thus final mixture of the two gases is at temperature 360 K. If final rms angular velocity of diatomic gas molecules is ω rms fâ¡ , according to law of equipartition of energy, we have
1 2 I ω rms fâ¡ 2 = kT
or ω rms fâ¡ = 2 kT I
= 2 × 1.38 × 1 0 - 2 3 × 3 6 0 2.76 × 1 0 - 4 6
ω rms fâ¡ = 6 × 1 0 1 2 rad/s

Topic: Physics, Kinetic Theory of Gases, JEE Main
An advantage of excreting nitrogenous wastes in the form of uric acid is that:

Options:
A Uric acid can be excreted in almost solid form.
B The formation of uric acid requires a least amount of energy.
C Uric acid is the first metabolic breakdown product of acids
D Uric acid may be excreted through the lungs

The correct answer is:
Uric acid can be excreted in almost solid form.

Solution:
The animals which excrete mainly uric acid are uricotelic. Uric acid can be stored or excreted in crystalline form.

Topic: Biology, Excretory Products and their Elimination, NEET
An aeroplane flying at a constant speed, parallel to the horizontal ground, $\sqrt{3} \mathrm{~km}$ above it, is observed at an elevation of $60^{\circ}$ from a point on the ground. If, after five seconds, its elevation from the same point, is $30^{\circ}$, then the speed (in $\mathrm{km} / \mathrm{hr}$ ) of the aeroplane is

Options:
A
1500

B
750

C
720

D
1440

The correct answer is:
1440

Solution:
For $\Delta \mathrm{OA}, \mathrm{A}, \mathrm{OA}_1=\frac{\sqrt{3}}{\tan 60^{\circ}}=1 \mathrm{~km}$
For $\Delta \mathrm{OB}_1, \mathrm{~B}, \mathrm{OB}_1=\frac{\sqrt{3}}{\tan 30^{\circ}}=3 \mathrm{~km}$.
As, a distance of $3-1=2 \mathrm{~km}$ is covered in 5 seconds. Therefore the speed of the plane is
$$
\frac{2 \times 3600}{5}=1440 \mathrm{~km} / \mathrm{hr}
$$

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
An aeroplane flying at a constant speed, parallel to the horizontal ground, 3 km above it is observed at an elevation of 60° from a point on the ground. If after five seconds, its elevation from the same point is 30°, then the speed (in km/hr) of the aeroplane is

Options:
A 720
B 1500
C 750
D 1440

The correct answer is:
1440

Solution:
Let from point C the angle of elevation of plane at B is 60°.

and after 5 seconds it reach at B'

In ∆ABC, AC=3cot60°=1

In ΔCA'B', A'C=3 cot30°=3

Hence, distance AA'=A'C-AC=3-1=2

Now, speed equals total distance travelled divided by total time taken.

Thus, speed =2560×60=2×60×605=72005=1440 km/hr

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2018 (15 Apr)
An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60o and after 10 seconds the elevation is observed to be 30o. The uniform speed of the aeroplane (in km/h), is

Options:
A 240
B 2403
C 603
D None of these

The correct answer is:
2403

Solution:
Let, d is the distance covered by aeroplane in 10 sec.
Let, the height at which the plane is flying be H=1 km
⇒H=AC=BD=1 km
In ∆OAC,
tan⁡60o=ACOC
OC=1cot⁡60o ...(i)
In ∆OBD,
tan⁡30o=BDOD
OD=1cot⁡30o ...(ii)
Now,
d=AB=CD=OD-OC
d=cot⁡30o-cot⁡60o
Speed=dt=cot⁡30o-cot⁡60o10km/sec
Speed in km/h
=cot⁡30o-cot⁡60o10×60×60
=3-1310×3600
=2403km/h

Topic: Mathematics, Heights and Distances, JEE Main
An aeroplane is flying at a velocity of 900 km h-1 loops a vertical circular loop. If the maximum force pressing the pilot against the seat is five times his weight, what would be the diameter (in m) of the loop? [ g=10 m s-2 ]

Options:

The correct answer is:
3125

Solution:
v=900km h-1
= 900 Ã 5 18
=250 m s-1
the maximum force is at the bottom of the vertical circle
Fmax=mv2r+mg=5mg
v2=4gr
r=v24g=2502410=1562.5 m

d=2r=3125 m

Topic: Physics, Laws of Motion, JEE Main
An aeroplane moving horizontally with a speed of $720 \mathrm{~km} / \mathrm{h}$ drops a food pocket, while flying at a height of $396.9 \mathrm{~m}$. the time taken by a food pocket to reach the ground and its horizontal range is (Take $g=9.8 \mathrm{~m} / \mathrm{sec}^2$ )

Options:
A $3 \mathrm{sec}$ and $2000 \mathrm{~m}$
B $5 \mathrm{sec}$ and $500 \mathrm{~m}$
C $8 \mathrm{sec}$ and $1500 \mathrm{~m}$
D $9 \mathrm{sec}$ and $1800 \mathrm{~m}$

The correct answer is:
$9 \mathrm{sec}$ and $1800 \mathrm{~m}$

Solution:
$t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 396.9}{9.8}} \simeq 9 \mathrm{sec}$ and $u=720 \mathrm{~km} / \mathrm{hr}=200 \mathrm{~m} / \mathrm{s}$
$\therefore R=u \times t=200 \times 9=1800 \mathrm{~m}$

Topic: Physics, Motion In Two Dimensions, JEE Main
An air bubble in a glass slab with refractive index 1.5 (near-normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness (in cm) of the slab is

Options:
A 8
B 10
C 12
D 16

The correct answer is:
12

Solution:
Apparent depth =Real depth from surfaceRefractive index
d=d1+d2μ
=1.55+3=12 cm

Topic: Physics, Ray Optics, JEE Main
An air bubble of diameter 6 mm rises steadily through a solution of density 1750 kg m-3 at the rate of 0.35 cm s-1. The co-efficient of viscosity of the solution (neglect density of air) is ________ Pas (given, g =10 m s-2).

Options:

The correct answer is:
10

Solution:
The bubble is moving with a constant velocity so the net force acting on the body is zero.

Stokes' law defines the drag force which is given by F=6πηrv

The buoyant force is given by B=Vρg=43πr3ρg

Hence, it can be written that

43πr3ρg=6πηrv⇒η=2r2ρg9v=2×3×10-32×1750×109×0.35×10-2=10 Pas

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2023 (08 Apr Shift 1)