67486 + Interview Questions in Secondary School in Physics

1.	What is stopping potential
Answer» ANSWER: The MINIMUM retarding means the negative potential of the ANODE of a photoelectric tube for which photoelectric CURRENT stops or becomes zero is called stopping potential.

1.

What is stopping potential

Answer»

The MINIMUM retarding means the negative potential of the ANODE of a photoelectric tube for which photoelectric CURRENT stops or becomes zero is called stopping potential.

Discussion

2.	Only for 12 class !!☺● Derive mirror formula Using Convex and concave mirror .Detailed explanation neededwith figure !!Do not Copy paste ❎❎
Answer» HEYA!! CHECK out the above ATTACHMENT

2.

Only for 12 class !!☺● Derive mirror formula Using Convex and concave mirror .Detailed explanation neededwith figure !!Do not Copy paste ❎❎

Answer»

HEYA!!

CHECK out the above ATTACHMENT

Discussion

3.	The photoelectric WOrK Function for a metalsurface 2.4ev if the light of wavelength 5000 A.is incident on the surface of metal find thresholdfeequency & incident frequency will there be emmision-OF photoelectrons or not.
Answer» ANSWER: REFER to the PHOTO. Explanation: refer to the photo.

3.

The photoelectric WOrK Function for a metalsurface 2.4ev if the light of wavelength 5000 A.is incident on the surface of metal find thresholdfeequency & incident frequency will there be emmision-OF photoelectrons or not.

Answer»

ANSWER:

REFER to the PHOTO.

Explanation:

refer to the photo.

Discussion

4.	a proton carrying 1MeV kinetic energy is moving in a circular path or radius r in uniform magnetic field . what should be the energy of alpha particle to describe a circle of same radius in the field
Answer» ANSWER: For a particle of mass m and charge Q in a magnetic FIELD, the radius of the path is given by; R= $\sqrt{x}$ *2M(KE)/qB* q∝ $\sqrt{x}$ *m(KE)* *2e/e=* $\sqrt{x}$ *(mp)(IMeV)/(4m p)(KE)* *1/4= 1/4(KE)* *KE=1MeV*

4.

a proton carrying 1MeV kinetic energy is moving in a circular path or radius r in uniform magnetic field . what should be the energy of alpha particle to describe a circle of same radius in the field

Answer»

ANSWER:

For a particle of mass m and charge Q in a magnetic FIELD, the radius of the path is given by;

R= $\sqrt{x}$ 2M(KE)/qB

q∝ $\sqrt{x}$ m(KE)

2e/e= $\sqrt{x}$ (mp)(IMeV)/(4m p)(KE)

1/4= 1/4(KE)

KE=1MeV

Discussion

5.	Define the term power of accomdation?
Answer» ANSWER: the ABILITY of EYE lens to adjust it's focal length☺️

5.

Define the term power of accomdation?

Answer»

ANSWER:

the ABILITY of EYE lens to adjust it's focal length☺️

Discussion

6.	A particle is given an initial velocity of (i + 2ſ)ms, g = 10 ms?, the equation of trajectory is ay = x - 5x2b) y = 2x - 5x2 c) 4y = 2x - 5x?d) 4y = 2x - 25x2
Answer» Question :- A particle is given an initial velocity of $\sf \hat{i} + 2 \hat{j} \: \frac{m}{s} \: ,g = 10 \frac{m}{s}$ ,the equation of TRAJECTORY is - (A) y = x - 5 x² (B) y = 2x - 5x² (C) 4Y = 2x - 5x² (D) 4y = 2x - 25x² Answer :- $\implies \boxed{\sf \: y = 2x \: - 5 {x}^{2} } \:$ → OPTION (B) is correct . To find :- → Equation of trajectory Explanation :- Given that :- $\to \sf initial \: velocity \: (u) = (\sf \hat{i} + 2 \hat{j} )\: \frac{m}{s} \: \\ \\ \star \: \sf \: vertical \: component \: of \: initial \: velocity \:i s \\ \: \sf \: \: u_y = 2 \: \frac{m}{s} \: \\ \\ \star \sf \:horizontal\: component \: of \: initial \: velocity \:i s \\ \: \sf \: \: u_x = 1 \: \frac{m}{s} \: \:$ Hence,apply equation of motion for horizontal component ( x - axis ) $\implies \boxed{\sf \: s = ut + \frac{1}{2} a {t}^{2} } \\ \therefore \\ \\ \implies \sf \: x = u_x \: t - \frac{1}{2} a \: {t}^{2} \: \: \: \: \: \: \: \: \because \: \: a_x = 0 \: \\ \\ \implies \sf x = u_x \: t \: \: \: \: \: \: \: \: \because \: u_x = 1 \\ \\ \to \boxed{ \sf x = t}$ Now, apply SECOND equation of motion for vertical component - $\to \sf y = u_y \: t + \frac{1}{2} a_y \: {t}^{2} \\ \\ \because \sf u_y= 2 \: \: and \: a_y = - g = - 10 \\ \\ \to \sf y = 2t \: - \frac{1}{2} \times 10 \times {t}^{2} \\ \\ \to \sf \: y = 2t \: - 5 {t}^{2} \\ \\ \because \sf \: x = t \\ \\ \implies \boxed{\sf \: y = 2x \: - 5 {x}^{2} }$

6.

A particle is given an initial velocity of (i + 2ſ)ms, g = 10 ms?, the equation of trajectory is ay = x - 5x2b) y = 2x - 5x2 c) 4y = 2x - 5x?d) 4y = 2x - 25x2

Answer»

Question :-

A particle is given an initial velocity of $\sf \hat{i} + 2 \hat{j} \: \frac{m}{s} \: ,g = 10 \frac{m}{s}$ ,the equation of TRAJECTORY is -

(A) y = x - 5 x²

(B) y = 2x - 5x²

(C) 4Y = 2x - 5x²

(D) 4y = 2x - 25x²

Answer :-

$\implies \boxed{\sf \: y = 2x \: - 5 {x}^{2} } \:$

→ OPTION (B) is correct .

To find :-

→ Equation of trajectory

Explanation :-

Given that :-

$\to \sf initial \: velocity \: (u) = (\sf \hat{i} + 2 \hat{j} )\: \frac{m}{s} \: \\ \\ \star \: \sf \: vertical \: component \: of \: initial \: velocity \:i s \\ \: \sf \: \: u_y = 2 \: \frac{m}{s} \: \\ \\ \star \sf \:horizontal\: component \: of \: initial \: velocity \:i s \\ \: \sf \: \: u_x = 1 \: \frac{m}{s} \: \:$

Hence,apply equation of motion for horizontal component ( x - axis )

$\implies \boxed{\sf \: s = ut + \frac{1}{2} a {t}^{2} } \\ \therefore \\ \\ \implies \sf \: x = u_x \: t - \frac{1}{2} a \: {t}^{2} \: \: \: \: \: \: \: \: \because \: \: a_x = 0 \: \\ \\ \implies \sf x = u_x \: t \: \: \: \: \: \: \: \: \because \: u_x = 1 \\ \\ \to \boxed{ \sf x = t}$

Now, apply SECOND equation of motion for vertical component -

$\to \sf y = u_y \: t + \frac{1}{2} a_y \: {t}^{2} \\ \\ \because \sf u_y= 2 \: \: and \: a_y = - g = - 10 \\ \\ \to \sf y = 2t \: - \frac{1}{2} \times 10 \times {t}^{2} \\ \\ \to \sf \: y = 2t \: - 5 {t}^{2} \\ \\ \because \sf \: x = t \\ \\ \implies \boxed{\sf \: y = 2x \: - 5 {x}^{2} }$

Discussion

7.	An object is kept at 15 cm from a convex mirror having radius of curvature 30cm. Theposition of image is formed at a distance of ......(a) 45cm(b) 15cm (c) 7.5 cm (d) 30cm
Answer» ANSWER: Your answer is 7.5 cm. hope it HELPS you. plz MARK me as BRANLIEST

7.

An object is kept at 15 cm from a convex mirror having radius of curvature 30cm. Theposition of image is formed at a distance of ......(a) 45cm(b) 15cm (c) 7.5 cm (d) 30cm

Answer»

ANSWER:

Your answer is 7.5 cm. hope it HELPS you. plz MARK me as BRANLIEST

Discussion

8.	A kingfisher flying at a height of 5m above the clear water lake sees a fish under water.The fish appears to be 3cm below the water surface. How much deep should the kingfisherimmerse its beak to pick that fish?(a) 4cm(b) 4m(C) 2.25 cm (d) 2.25m
Answer» Answer: 2.5 cm Explanation: jjjjjjjjjjjjjjjjjjjjjjjj

8.

A kingfisher flying at a height of 5m above the clear water lake sees a fish under water.The fish appears to be 3cm below the water surface. How much deep should the kingfisherimmerse its beak to pick that fish?(a) 4cm(b) 4m(C) 2.25 cm (d) 2.25m

Answer»

Answer:

2.5 cm

Explanation:

jjjjjjjjjjjjjjjjjjjjjjjj

Discussion

9.	What is the function of galvanometer in a circuit?
Answer» ANSWER: it MEASURES the CURRENT and VOLTAGE

9.

What is the function of galvanometer in a circuit?

Answer»

ANSWER:

it MEASURES the CURRENT and VOLTAGE

Discussion

10.	(v)The equation of trajectory of a projectile is in the form of ....(a) hyperbola(b) ellipse(c) both (a) and (b)(d) parabola
Answer» Answer: i THINK PARABOLA is UR answerrrr Explanation: HÔPË ÏTẞ HÉLPFŪLL PLS MARK AS BRAINLIEST

10.

(v)The equation of trajectory of a projectile is in the form of ....(a) hyperbola(b) ellipse(c) both (a) and (b)(d) parabola

Answer»

Answer:

i THINK PARABOLA is UR answerrrr

Explanation:

HÔPË ÏTẞ HÉLPFŪLL

PLS MARK AS BRAINLIEST

Discussion

11.	Which of the following is not a state function?a) Work done at constant pressureb) Enthalpyc) Work done by conservative forced) Work done by non-conservative force
Answer» ANSWER: WORK done by non CONSERVATIVE force

11.

Which of the following is not a state function?a) Work done at constant pressureb) Enthalpyc) Work done by conservative forced) Work done by non-conservative force

Answer»

ANSWER:

WORK done by non CONSERVATIVE force

Discussion

12.	(iv)The two unidirectional vectors P andhaving same length are called.....(a) negative vectors(b) equal vectors(©) null vectors(d) unit vectors
Answer» ANSWER: NULL VECTOR MAY be the RIGHT answer

12.

(iv)The two unidirectional vectors P andhaving same length are called.....(a) negative vectors(b) equal vectors(©) null vectors(d) unit vectors

Answer»

ANSWER:

NULL VECTOR MAY be the RIGHT answer

Discussion

13.	Converging or diverging ability of a lens is called it's .....(a) focal power(b) dispersive power(c) magnifying power(d) refractive index
Answer» Answer: focal POWER The converging or DIVERGING ability is called as the power of LENS.

13.

Converging or diverging ability of a lens is called it's .....(a) focal power(b) dispersive power(c) magnifying power(d) refractive index

Answer»

Answer: focal POWER

The converging or DIVERGING ability is called as the power of LENS.

Discussion

14.	Find force when mass of object is increased 4 times
Answer» Answer: hlo mate.. here's UR answer... F=ma i.e force is DIRECTLY proportional to MASS THEREFORE force will also BECOME 4times.. hope it helps..

14.

Find force when mass of object is increased 4 times

Answer»

Answer:

hlo mate..

here's UR answer...

F=ma

i.e force is DIRECTLY proportional to MASS THEREFORE force will also BECOME 4times..

hope it helps..

Discussion

15.	State four laws of photo electric emission.
Answer» Answer: 1. Instaneous process 2.Threshold frequency 3. photoeletric current is DIRECTLY proportional to the INTENSITY of INCIDENT light 4.Maximum VELOCITY of the electrons depends upon the frequency of incident light. Explanation:

15.

State four laws of photo electric emission.

Answer»

Answer:

1. Instaneous process

2.Threshold frequency

3. photoeletric current is DIRECTLY proportional to the INTENSITY of INCIDENT light

4.Maximum VELOCITY of the electrons depends upon the frequency of incident light.

Explanation:

Discussion

16.	12. Write four properties of X-rays.13. What are the conditions fon the tun1:11
Answer» Answer: 1.they are capable of travelling in vaccum 2.requires HIGH voltage to produce xray 3.they travel in STRAIGHT line and do not carry any electric charge 4.tgey are used to capture HUMAN SKELETON defects

16.

12. Write four properties of X-rays.13. What are the conditions fon the tun1:11

Answer»

Answer:

1.they are capable of travelling in vaccum

2.requires HIGH voltage to produce xray

3.they travel in STRAIGHT line and do not carry any electric charge

4.tgey are used to capture HUMAN SKELETON defects

Discussion

17.	A uniform meter rod of 2 N is pivoted at 60 cm mark.The total length of meter rod is 100 cm.If a load of 4 N is suspended at one end 40 cm from pivot point.What is the net torque about the pivot.
Answer» Torque DUE to LOAD =4N×40cm=1.6Nm Torque due to WEIGHT of the RULE =2N×10cm=0.2Nm Both are in opposite to each other so NET torque =1.6−0.2=1.4Nm

17.

A uniform meter rod of 2 N is pivoted at 60 cm mark.The total length of meter rod is 100 cm.If a load of 4 N is suspended at one end 40 cm from pivot point.What is the net torque about the pivot.

Answer»

Torque DUE to LOAD =4N×40cm=1.6Nm

Torque due to WEIGHT of the RULE =2N×10cm=0.2Nm

Both are in opposite to each other so NET torque =1.6−0.2=1.4Nm

Discussion

18.	14, The fringe width in a Young's double slit interference pattern is2.4x104m, when red light of wavelength 64008 is used. By howmuch will it change, if blue light of wavelength 40008 is used ?
Answer» J h j j K k k k j that j j jj mk ms MO KB kn j

18.

14, The fringe width in a Young's double slit interference pattern is2.4x104m, when red light of wavelength 64008 is used. By howmuch will it change, if blue light of wavelength 40008 is used ?

Answer»

J h j j K k k k j that j j jj mk ms MO KB kn j

Discussion

19.	Why is it that while using a movingcoil galvanometer as a voltmeter, ahigh resistance in series is requiredwhereas in an ammeter a shunt isused? All India 2010; Delhi 2009
Answer» Answer: 111111111111111111111111111

19.

Why is it that while using a movingcoil galvanometer as a voltmeter, ahigh resistance in series is requiredwhereas in an ammeter a shunt isused? All India 2010; Delhi 2009

Answer»

Answer:

111111111111111111111111111

Discussion

20.	Q5. two springs of constants k1 and k2 have equal maximum velocities whenexecuting SHM. The ratio of their amplitudes will be (masses are equal)-
Answer» Question : Two SPRINGS of constants k1 and k2 have equal maximum VELOCITIES when executing SHM,the ratio of their amplitudes will be (MASSES are equal) : Solution : The springs have equal maximum velocities and masses Their springs constants are $\sf K_1 \ \sf{and} \ K_2$ Maximum Velocity is GIVEN by : $\sf \: \: v {}_{max} = A \omega$ Also, $\sf \: \omega \: = \sqrt{ \dfrac{k}{m} }$ THUS, $\sf \: v = {A}_{1} \times \sqrt{ \dfrac{ K_{1} }{M} } - - - - - - - (1)$ Also, $\sf \: v = A_{2} \times \sqrt{ \dfrac{ K_{2} }{M} } - - - - - - -(2)$ Equating equations (1) and (2),as maximum velocities of the SHM are equal $\longrightarrow \: \sf \: A_{1} \times \sqrt{ \dfrac{ K_{1} }{M} } = A_{2} \times \sqrt{ \dfrac{ {K}_{2} }{M} } \\ \\ \longrightarrow \: \sf{ \dfrac{ A_{1} }{ A_{2} } = \sqrt{ \dfrac{ K_{2} }{ K_{1} } } } \\ \\ \longrightarrow \: \boxed{ \boxed{\sf{ \dfrac{ A_{1} }{ A_{2} } = (\frac{ K_{1} }{ {K}_{2} } ) {}^{ \frac{1}{2} } }}}$

20.

Q5. two springs of constants k1 and k2 have equal maximum velocities whenexecuting SHM. The ratio of their amplitudes will be (masses are equal)-

Answer»

Question :

Two SPRINGS of constants k1 and k2 have equal maximum VELOCITIES when executing SHM,the ratio of their amplitudes will be (MASSES are equal) :

Solution :

The springs have equal maximum velocities and masses

Their springs constants are $\sf K_1 \ \sf{and} \ K_2$

Maximum Velocity is GIVEN by :

$\sf \: \: v {}_{max} = A \omega$

Also,

$\sf \: \omega \: = \sqrt{ \dfrac{k}{m} }$

THUS,

$\sf \: v = {A}_{1} \times \sqrt{ \dfrac{ K_{1} }{M} } - - - - - - - (1)$

Also,

$\sf \: v = A_{2} \times \sqrt{ \dfrac{ K_{2} }{M} } - - - - - - -(2)$

Equating equations (1) and (2),as maximum velocities of the SHM are equal

$\longrightarrow \: \sf \: A_{1} \times \sqrt{ \dfrac{ K_{1} }{M} } = A_{2} \times \sqrt{ \dfrac{ {K}_{2} }{M} } \\ \\ \longrightarrow \: \sf{ \dfrac{ A_{1} }{ A_{2} } = \sqrt{ \dfrac{ K_{2} }{ K_{1} } } } \\ \\ \longrightarrow \: \boxed{ \boxed{\sf{ \dfrac{ A_{1} }{ A_{2} } = (\frac{ K_{1} }{ {K}_{2} } ) {}^{ \frac{1}{2} } }}}$

Discussion

21.	Ii) Write the sulphate ore of magnesium ?
Answer» ANSWER: ouccrs in SEA WATER, mineral SPRINGS and in MINERALS such as kieserite and epsomite Explanation:

21.

Ii) Write the sulphate ore of magnesium ?

Answer»

ANSWER:

ouccrs in SEA WATER, mineral SPRINGS and in MINERALS such as kieserite and epsomite

Explanation:

Discussion

22.	Define angular displacement, angular velocity & angularQueleration
Answer» ANSWER: Angular DISPLACEMENT : The angle described the particle about the AXIS of ROTATION( or center O) in a given TIME. Angular velocity : The rate of change of angular displacement .

22.

Define angular displacement, angular velocity & angularQueleration

Answer»

ANSWER:

Angular DISPLACEMENT :

The angle described the particle about the AXIS of ROTATION( or center O)

in a given TIME.

Angular velocity :

The rate of change of angular displacement .

Discussion

23.	3.Find the effective resistance between the points A and B in figure.
Answer» ANSWER: there is no FIGURE Explanation: where is figure

23.

3.Find the effective resistance between the points A and B in figure.

Answer»

ANSWER:

there is no FIGURE

Explanation:

where is figure

Discussion

24.	Plz answer this question(A) Device X is a concave mirror and device Y is a convex lens, whosefocal lengths are 20 cm and 25 cm respectively."(B) Device X is a convex lens and device Y is a concave mirror, whose focallengths are 10 cm and 25 cm respectively.(C) Device X is a concave lens and device Y is a convex mirror, whose focallengths are 20 cm and 25 cm respectively.(D) Device X is a convex lens and device Y is a concave mirror, whose focallengths are 20 cm and 25 cm respectively.(CBSE 2017)
Answer» EXPLANATION: the OPTION a is APPROPRIATE

24.

Plz answer this question(A) Device X is a concave mirror and device Y is a convex lens, whosefocal lengths are 20 cm and 25 cm respectively."(B) Device X is a convex lens and device Y is a concave mirror, whose focallengths are 10 cm and 25 cm respectively.(C) Device X is a concave lens and device Y is a convex mirror, whose focallengths are 20 cm and 25 cm respectively.(D) Device X is a convex lens and device Y is a concave mirror, whose focallengths are 20 cm and 25 cm respectively.(CBSE 2017)

Answer»

EXPLANATION:

the OPTION a is APPROPRIATE

Discussion

25.	IFind the effective resistance between the points A and B in figure.[2]sa.V
Answer» ANSWER: The effective resistance in series combination of two equal resistance is ′s′. When they are JOINED in parallel the total resistance is p. If s=np then the minimum POSSIBLE VALUE of ′n′ is.

25.

IFind the effective resistance between the points A and B in figure.[2]sa.V

Answer»

ANSWER:

The effective resistance in series combination of two equal resistance is ′s′. When they are JOINED in parallel the total resistance is p. If s=np then the minimum POSSIBLE VALUE of ′n′ is.

Discussion

26.	A force of magnitude 10 units makes an angle of 30° with x axis. The horizontal andvertical component of given force are(a) 5v3 unit, 5 unit (b) 5 unit, 5v3 unit (c) 10v3 unit, 10 unit (d) 10 unit, 10v3 unit
Answer» Answer: B) 5√3 unit, 5 unit Explanation: Horizontal component of the GIVEN FORCE = Fcos3O° Vertical component of given force = Fsin3O° where F = \|O(TEN) unit (given)

26.

A force of magnitude 10 units makes an angle of 30° with x axis. The horizontal andvertical component of given force are(a) 5v3 unit, 5 unit (b) 5 unit, 5v3 unit (c) 10v3 unit, 10 unit (d) 10 unit, 10v3 unit

Answer»

Answer:

B) 5√3 unit, 5 unit

Explanation:

Horizontal component of the GIVEN FORCE

= Fcos3O°

Vertical component of given force

= Fsin3O°

where F = |O(TEN) unit (given)

Discussion

27.	. 04. The RMS velocity of Nitrogen molecules at N.T.P. is?
Answer» ANSWER: The R.M.S velocity of Nitrogen molecules at N.T.P. is 493 m/s Given: Here, N.T.P specifies STANDARD conditions. Therefore, Temperature will be 273 K. Molar mass of Nitrogen = 28 grams. Explanation: $\rule{300}{1.5}$ From the Formula, $\large \bigstar\;{\boxed{\tt V_{RMS} = \sqrt{\dfrac{3\;R\;T\;}{M}}}}$ $\bold{Here}\begin{cases}\text{R Denotes Universal gas constant} \\ \text{T Denotes Temperature} \\ \text{M Denotes Molar mass (Kg)} \end{cases}$ Now, $\large{\boxed{\tt V_{RMS} = \sqrt{\dfrac{3\;R\;T}{M}}}}$ Substituting the VALUES, $\longmapsto\large{\tt V_{RMS} = \sqrt{\dfrac{3 \times 8.314 \times 273}{0.028}}}$ As, R = 8.314 J/mol.K Mass of Nitrogen = 28 grams = 0.028 Kg. $\longmapsto\large{\tt V_{RMS} = \sqrt{\dfrac{24.942 \times 273}{0.028}}}$ $\longmapsto\large{\tt V_{RMS} = \sqrt{\dfrac{6809.166}{0.028}}}$ $\longmapsto\large{\tt V_{RMS} = \sqrt{243184.5}}$ $\longmapsto\large{\underline{\boxed{\red{\tt V_{RMS} = 493.137 \; m/s}}}}$ ∴ The R.M.S velocity of Nitrogen molecules at N.T.P. is 493 m/s (approx). $\boxed{\begin{minipage}{7.5cm} $ \bigstar \underline{\text{Extra Formulas}} \\\\ \tt \star \;V_{(average)} = \sqrt{\dfrac{8 \; R \; T}{\pi \; M}} \\\\ \tt \star \;V_{(mp)} = \sqrt{\dfrac{2 \; R \; T}{ M}} \\\\ \star \; \text{Ratio of Three velocities,} \\ \longmapsto V_{(RMS)} : V_{(mp)} : V_{(Average)} = \sqrt{3} : \sqrt{2} : \sqrt{\dfrac{8}{\pi}} \\\\ \rule{200}{1} \\ \text{Note} \\ \bullet\; V_{(mp)} = \text{Most probable Speed} \\ \bullet\;V_{(average)} = \text{Average Speed} $ \end{minipage}}$ $\rule{300}{1.5}$

27.

. 04. The RMS velocity of Nitrogen molecules at N.T.P. is?

Answer»

ANSWER:

The R.M.S velocity of Nitrogen molecules at N.T.P. is 493 m/s

Given:

Here, N.T.P specifies STANDARD conditions.

Therefore, Temperature will be 273 K.
Molar mass of Nitrogen = 28 grams.

Explanation:

$\rule{300}{1.5}$

From the Formula,

$\large \bigstar\;{\boxed{\tt V_{RMS} = \sqrt{\dfrac{3\;R\;T\;}{M}}}}$

$\bold{Here}\begin{cases}\text{R Denotes Universal gas constant} \\ \text{T Denotes Temperature} \\ \text{M Denotes Molar mass (Kg)} \end{cases}$

Now,

$\large{\boxed{\tt V_{RMS} = \sqrt{\dfrac{3\;R\;T}{M}}}}$

Substituting the VALUES,

$\longmapsto\large{\tt V_{RMS} = \sqrt{\dfrac{3 \times 8.314 \times 273}{0.028}}}$

As,

R = 8.314 J/mol.K
Mass of Nitrogen = 28 grams = 0.028 Kg.

$\longmapsto\large{\tt V_{RMS} = \sqrt{\dfrac{24.942 \times 273}{0.028}}}$

$\longmapsto\large{\tt V_{RMS} = \sqrt{\dfrac{6809.166}{0.028}}}$

$\longmapsto\large{\tt V_{RMS} = \sqrt{243184.5}}$

$\longmapsto\large{\underline{\boxed{\red{\tt V_{RMS} = 493.137 \; m/s}}}}$

∴ The R.M.S velocity of Nitrogen molecules at N.T.P. is 493 m/s (approx).

$\boxed{\begin{minipage}{7.5cm} $ \bigstar \underline{\text{Extra Formulas}} \\\\ \tt \star \;V_{(average)} = \sqrt{\dfrac{8 \; R \; T}{\pi \; M}} \\\\ \tt \star \;V_{(mp)} = \sqrt{\dfrac{2 \; R \; T}{ M}} \\\\ \star \; \text{Ratio of Three velocities,} \\ \longmapsto V_{(RMS)} : V_{(mp)} : V_{(Average)} = \sqrt{3} : \sqrt{2} : \sqrt{\dfrac{8}{\pi}} \\\\ \rule{200}{1} \\ \text{Note} \\ \bullet\; V_{(mp)} = \text{Most probable Speed} \\ \bullet\;V_{(average)} = \text{Average Speed} $ \end{minipage}}$

$\rule{300}{1.5}$

Discussion

28.	(b) A projectile with initial speed u, can have the same range R for two angles ofprojection. If t1 and t2 be the time of flight in the two cases, then find the value ofT1T2 in terms of R
Answer» Answer: GIVEN: A PROJECTILE with speed U , has same range R , for 2 different time period. To find: t1 × t2 = ? (in terms of R) Calculation: For constant Projection velocity and same range , this condition is only possible only when a Projectile has been thrown at complementary angles. Let one angle be θ and the other be (90 - θ). Calculation: Time for 1st Projection be t1 $t1 = \dfrac{2u \sin( \theta) }{g}$ Time for 2nd Projection be t2 $t2 = \dfrac{2u \sin(90 \degree - \theta) }{g}$ $= > t2 = \dfrac{2u \cos( \theta) }{g}$ Now as per QUESTION : $\therefore \: t1 \times t2$ $= > \dfrac{2u \sin( \theta) }{g} \times \dfrac{2u \cos( \theta) }{g} \:$ $= > ( \dfrac{2}{g}) \times \dfrac{ {u}^{2} \times 2 \sin( \theta) \cos( \theta) }{g}$ $= > ( \dfrac{2}{g} ) \times \dfrac{ {u}^{2} \times \sin(2 \theta) }{g}$ $= > \dfrac{2R}{g}$ So final answer is: $\boxed{ \large{ \red{t1 \times t2 = \frac{2R}{g}}}}$

28.

(b) A projectile with initial speed u, can have the same range R for two angles ofprojection. If t1 and t2 be the time of flight in the two cases, then find the value ofT1T2 in terms of R

Answer»

Answer:

GIVEN:

A PROJECTILE with speed U , has same range R , for 2 different time period.

To find:

t1 × t2 = ? (in terms of R)

Calculation:

For constant Projection velocity and same range , this condition is only possible only when a Projectile has been thrown at complementary angles.

Let one angle be θ and the

other be (90 - θ).

Calculation:

Time for 1st Projection be t1

$t1 = \dfrac{2u \sin( \theta) }{g}$

Time for 2nd Projection be t2

$t2 = \dfrac{2u \sin(90 \degree - \theta) }{g}$

$= > t2 = \dfrac{2u \cos( \theta) }{g}$

Now as per QUESTION :

$\therefore \: t1 \times t2$

$= > \dfrac{2u \sin( \theta) }{g} \times \dfrac{2u \cos( \theta) }{g} \:$

$= > ( \dfrac{2}{g}) \times \dfrac{ {u}^{2} \times 2 \sin( \theta) \cos( \theta) }{g}$

$= > ( \dfrac{2}{g} ) \times \dfrac{ {u}^{2} \times \sin(2 \theta) }{g}$

$= > \dfrac{2R}{g}$

So final answer is:

$\boxed{ \large{ \red{t1 \times t2 = \frac{2R}{g}}}}$

Discussion

29.	15. () An electric kettle has two heating coils. When one of the coil is switched on theliquid in the kettle begins to boil in 15 minutes and when the other is switchedon the boiling begins in 18 minutes. In what time will the boiling begin if bothcoils are switched on simultaneously i) in series & (ii) in parallel.
Answer» Answer: in series the time of boiling will increase, in parallel the time which required will be decrease because the HEATING coil has its own INTERNAL resistance and when they are connected in parallel the resistance will decrease then the CURRENT FLOW will increase.

29.

15. () An electric kettle has two heating coils. When one of the coil is switched on theliquid in the kettle begins to boil in 15 minutes and when the other is switchedon the boiling begins in 18 minutes. In what time will the boiling begin if bothcoils are switched on simultaneously i) in series & (ii) in parallel.

Answer»

Answer:

in series the time of boiling will increase, in parallel the time which required will be decrease because the HEATING coil has its own INTERNAL resistance and when they are connected in parallel the resistance will decrease then the CURRENT FLOW will increase.

Discussion

30.	OR(a) 12 cells each having the same e.m.f. are connected in aeries and are kept in aclosed box. Some of the cells are wrongly connected. This battery is connectedin aeries with an ammeter and two cells identical with the others. The currentis 3 ampere when the cells and battery aid each other and is 2 ampere whenthe cells and battery oppose each other. How many 20cells in the battery are wrongly connected?[3+2](6) Calculate the steady state current in the 2 ohm resistorshown in the circuit in the figure. The internal resistance of the battery is negligible
Answer» Answer: uske BAAD mein dusro ko dosh dene ka koi adhikar NAHI HAI unko the following observations of the following link to me that the solar CELLS might help me with any questions you may be a part of a WORKING holiday visa application for Explanation: mdm

30.

OR(a) 12 cells each having the same e.m.f. are connected in aeries and are kept in aclosed box. Some of the cells are wrongly connected. This battery is connectedin aeries with an ammeter and two cells identical with the others. The currentis 3 ampere when the cells and battery aid each other and is 2 ampere whenthe cells and battery oppose each other. How many 20cells in the battery are wrongly connected?[3+2](6) Calculate the steady state current in the 2 ohm resistorshown in the circuit in the figure. The internal resistance of the battery is negligible

Answer»

Answer:

uske BAAD mein dusro ko dosh dene ka koi adhikar NAHI HAI unko the following observations of the following link to me that the solar CELLS might help me with any questions you may be a part of a WORKING holiday visa application for

Explanation:

mdm

Discussion

31.	Body starts from rest moving withvelocity 144 km/her cover a distance,5m in 10 sec find acceleratforce and work done.find the cravitational force blog
Answer» INTIAL velocity=0 final velocity=144 time=10 sec=1/360 a=v-u/t =144-0/1/360 =144/1/360 =144360 =51840m/ss

31.

Body starts from rest moving withvelocity 144 km/her cover a distance,5m in 10 sec find acceleratforce and work done.find the cravitational force blog

Answer»

INTIAL velocity=0

final velocity=144

time=10 sec=1/360

a=v-u/t

=144-0/1/360

=144/1/360

=144*360

=51840m/s*s

Discussion

32.	A man stands in a lift that is accelerating verticallydownwards. Which statement describes the force exerted bythe man on the floor? A. It is equal to the weight of man.B)It is greater than the force exerted by floor on the manC.It is less than the force exerted by the floor on the man.D)It is less than the weight of man.
Answer» ANSWER: (c) is your CORRECT answer MARK it's BRAINLIEST and please follow me

32.

A man stands in a lift that is accelerating verticallydownwards. Which statement describes the force exerted bythe man on the floor? A. It is equal to the weight of man.B)It is greater than the force exerted by floor on the manC.It is less than the force exerted by the floor on the man.D)It is less than the weight of man.

Answer»

ANSWER:

(c) is your CORRECT answer

MARK it's BRAINLIEST and please follow me

Discussion

33.	Find the correlation between the terms giAngle of deviation and wavelength of light in one sentence
Answer» Answer: The ANGLE of minimum deviation is INVERSELY proportional to the wavelength of INCIDENT RAY. Hence , when the wavelength of the incident ray is REDUCED the angle of minimum deviation INCREASES mark the brainliest Explanation:

33.

Find the correlation between the terms giAngle of deviation and wavelength of light in one sentence

Answer»

Answer:

The ANGLE of minimum deviation is INVERSELY proportional to the wavelength of INCIDENT RAY.

Hence , when the wavelength of the incident ray is REDUCED the angle of minimum deviation INCREASES

mark the brainliest

Explanation:

Discussion

34.	9An element ‘X’has a Valency (3+):(a) Write the formula of its phosphate.(b) Write the formula of its carbonate.OR
Answer» Answer: GIVEN element X with valency 3 1)Phosphide means compound of X with P [phosphide valency 3] chemical FORMULA would be: X3P3 hence XP is the chemical formula of its phosphide compound. 2) Carbonate valency 2 chemical formula would be X2(CO)3 3) The element X should be metal because Phosphorous AND CARBON ARE nonmetals. *hope..it..helps..* *all..the..best..* *mark..me..brainliest...*

34.

9An element ‘X’has a Valency (3+):(a) Write the formula of its phosphate.(b) Write the formula of its carbonate.OR

Answer»

Answer:

GIVEN element X with valency 3

1)Phosphide means compound of X with P [phosphide valency 3]

chemical FORMULA would be:

X3P3

hence XP is the chemical formula of its phosphide compound.

2) Carbonate valency 2

chemical formula would be

X2(CO)3

3) The element X should be metal because Phosphorous AND CARBON ARE nonmetals.

hope..it..helps..

all..the..best..

mark..me..brainliest...

Discussion

35.	What is the name of Biggest super Cluster
Answer» Answer: he Hercules-Corona Borealis Great Wall. Explanation: It was FIRST reported in 2013 and has been studied several TIMES. It's so BIG that light takes about 10 billion YEARS to move across the structure.

35.

What is the name of Biggest super Cluster

Answer»

Answer:

he Hercules-Corona Borealis Great Wall.

Explanation:

It was FIRST reported in 2013 and has been studied several TIMES. It's so BIG that light takes about 10 billion YEARS to move across the structure.

Discussion

36.	Which acid is used in antacid
Answer» EXPLANATION: Antacid CONTAINS a BASE known as Magnesium Hydroxide.

36.

Which acid is used in antacid

Answer»

EXPLANATION:

Antacid CONTAINS a BASE known as Magnesium Hydroxide.

Discussion

37.	Only for class 12 !!Explain cause of reddish brown color of sun at sunset and sunrise ? Do not copy paste ❎_/\_
Answer» $\huge\underline\mathcal\green{Hii MatE}$ $\huge\underline\mathfrak\purple{Answer.!!!}$ The sun during sunrise and sunset is is NEAR the horizon and due to this when LIGHT comes they got scattered except red light which FALLS in our eyes and it sees reddish BROWN in COLOUR

37.

Only for class 12 !!Explain cause of reddish brown color of sun at sunset and sunrise ? Do not copy paste ❎_/\_

Answer»

$\huge\underline\mathcal\green{Hii MatE}$

$\huge\underline\mathfrak\purple{Answer.!!!}$

The sun during sunrise and sunset is is NEAR the horizon and due to this when LIGHT comes they got scattered except red light which FALLS in our eyes and it sees reddish BROWN in COLOUR

Discussion

38.	Darpan me u or v kya hota h
Answer» Answer: IMAGE DISTANCE from the pole is known as v nd object distance from the pole is known as u thank you

38.

Darpan me u or v kya hota h

Answer»

Answer:

IMAGE DISTANCE from the pole is known as v

nd object distance from the pole is known as u

thank you

Discussion

39.	Only for class 12 !!Explain the cause of Blue color of sky , on the basis of Scattering of light .Do not Copy paste ❎ Either it would be deleted !!
Answer» ANSWER: Blue light is scattered in all DIRECTIONS by the tiny MOLECULES of air in Earth's atmosphere. Blue is scattered more than other COLORS because it travels as shorter, smaller waves. This is why we see a blue sky most of the time. ... Also, the surface of Earth has reflected and scattered the light.

39.

Only for class 12 !!Explain the cause of Blue color of sky , on the basis of Scattering of light .Do not Copy paste ❎ Either it would be deleted !!

Answer»

ANSWER:

Blue light is scattered in all DIRECTIONS by the tiny MOLECULES of air in Earth's atmosphere. Blue is scattered more than other COLORS because it travels as shorter, smaller waves. This is why we see a blue sky most of the time. ... Also, the surface of Earth has reflected and scattered the light.

Discussion

40.	What is the answer of the following
Answer» ANSWER: RESISTANCE will not change Explanation: resistance of a WIRE does not DEPEND on it's LENGTH

40.

What is the answer of the following

Answer»

ANSWER:

RESISTANCE will not change

Explanation:

resistance of a WIRE does not DEPEND on it's LENGTH

Discussion

41.	What is the frequency of power in an AC circuit connected to a capacitor. Given 1 - 50 HzV = 220 V(1) 50 Hz(2) 100 H(3) zero(4) 60 Hzplzz answer this question...
Answer» Answer: The capacitor can't CHANGE the frequency of its SUPPLY. HENCE the Answer is A). 50 HZ.

41.

What is the frequency of power in an AC circuit connected to a capacitor. Given 1 - 50 HzV = 220 V(1) 50 Hz(2) 100 H(3) zero(4) 60 Hzplzz answer this question...

Answer»

Answer:

The capacitor can't CHANGE the frequency of its SUPPLY. HENCE the Answer is A). 50 HZ.

Discussion

42.	Derive the expression on maximum speed of circular motion of a car on a banked road
Answer» Consider the attachment; From the diagram, N= mgcosθ + Fc sinθ --(i) and, Fc cosθ = mgsinθ + μN (ii) COMBINING these two together by substituting the value, of N in the SECOND equation, Fc cosθ = mg sinθ + μ(mgcosθ + Fc sinθ) Solving FURTHUR, Fc cosθ - μFc sinθ= mg SIN θ + μmgcosθ Since Fc= mv²/r Solving furthur, $\frac{mv^{2} }{r} (cosθ -μsinθ)=mg(sinθ + μcosθ)\\\\\frac{v^{2} }{r} (cosθ -μsinθ) = g(sinθ + μcosθ)\\\\v^{2} =gr\frac{(sinθ + μcosθ)}{(cosθ -μsinθ)} \\\\v =\sqrt{gr\frac{(sinθ + μcosθ)}{(cosθ -μsinθ)}} \\$ Here, v = maximum velocity, r = radius of the banking, Fc= centripetal force, μ= coefficient of friction and N= normal reaction froce.

42.

Derive the expression on maximum speed of circular motion of a car on a banked road

Answer»

Consider the attachment;

From the diagram,

N= mgcosθ + Fc sinθ --(i)

and,

Fc cosθ = mgsinθ + μN (ii)

COMBINING these two together by substituting the value, of N in the SECOND equation,

Fc cosθ = mg sinθ + μ(mgcosθ + Fc sinθ)

Solving FURTHUR,

Fc cosθ - μFc sinθ= mg SIN θ + μmgcosθ

Since Fc= mv²/r

Solving furthur,

$\frac{mv^{2} }{r} (cosθ -μsinθ)=mg(sinθ + μcosθ)\\\\\frac{v^{2} }{r} (cosθ -μsinθ) = g(sinθ + μcosθ)\\\\v^{2} =gr\frac{(sinθ + μcosθ)}{(cosθ -μsinθ)} \\\\v =\sqrt{gr\frac{(sinθ + μcosθ)}{(cosθ -μsinθ)}} \\$

Here, v = maximum velocity, r = radius of the banking, Fc= centripetal force, μ= coefficient of friction and N= normal reaction froce.

Discussion

43.	Only for class 12 !! ✌Derive lens maker formula ??Do not spam ❎Either it would be deleted !!
Answer» HEYA!! CHECK out the above ATTACHMENT.

43.

Only for class 12 !! ✌Derive lens maker formula ??Do not spam ❎Either it would be deleted !!

Answer»

HEYA!!

CHECK out the above ATTACHMENT.

Discussion

44.	0.5 C of charge flows through a conductor in 0.5 seconds what is the magnitude of the current flowing through the conductor
Answer» ANSWER: Q= 0.5 t=0.5 I= Q/T I=0.5/0.5 I=1 AMP

44.

0.5 C of charge flows through a conductor in 0.5 seconds what is the magnitude of the current flowing through the conductor

Answer»

ANSWER:

Q= 0.5

t=0.5

I= Q/T

I=0.5/0.5

I=1 AMP

Discussion

45.	Using Biot-Savart’s law, derive an expression for magnetic field at any point onaxial line of a current carrying circular loop. Hence, find magnitude of magnetic field intensity at the centre of circular coil
Answer» Answer: Using Biot-Savart's LAW, derive an expression for MAGNETIC field at any point on axial LINE of a CURRENT carrying circular loop. Hence, find magnitude of magnetic field intensity at the centre of circular coil. And direction of →dB is ⊥ to the plane containing I→dl and →r. Resolving →dB alonhg the x-axis and y-axis.

45.

Using Biot-Savart’s law, derive an expression for magnetic field at any point onaxial line of a current carrying circular loop. Hence, find magnitude of magnetic field intensity at the centre of circular coil

Answer»

Answer:

Using Biot-Savart's LAW, derive an expression for MAGNETIC field at any point on axial LINE of a CURRENT carrying circular loop. Hence, find magnitude of magnetic field intensity at the centre of circular coil. And direction of →dB is ⊥ to the plane containing I→dl and →r. Resolving →dB alonhg the x-axis and y-axis.

Discussion

46.	Please help me with this guys....but a little early
Answer» graph i) is Correct answer as Constant SLOPE showing uniform acceleration Explanation: Uniform acceleration Means acceleration is constant => acceleration = k ( k is a constant) aceeleration = Velocity / Time a = dv/dt => dv/dt = k INTEGRATING both SIDES => v = kt + c c is another constant at t = 0 v = c so Slope is constant = k graph i) is Correct answer as in Graph ii) slope is not constant Learn more: Figure given below shows the graph of velocity v ofparticle MOVING ... brainly.in/question/10432386 Figure given below shows the graph of velocity v ofparticle moving ... brainly.in/question/11151179

46.

Please help me with this guys....but a little early

Answer»

graph i) is Correct answer as Constant SLOPE showing uniform acceleration

Explanation:

Uniform acceleration Means

acceleration is constant

=> acceleration = k ( k is a constant)

aceeleration = Velocity / Time

a = dv/dt

=> dv/dt = k

INTEGRATING both SIDES

=> v = kt + c

c is another constant

at t = 0 v = c

so

Slope is constant = k

graph i) is Correct answer

as in Graph ii) slope is not constant

Learn more:

Figure given below shows the graph of velocity v ofparticle MOVING ...

brainly.in/question/10432386

Figure given below shows the graph of velocity v ofparticle moving ...

brainly.in/question/11151179

Discussion

47.	An expression for time period of satellite orbiting around the earth of radius R and at heighth
Answer» EXPLANATION: S=v t ∵t=T T=S/v S=2πR T=2πR/v

47.

An expression for time period of satellite orbiting around the earth of radius R and at heighth

Answer»

EXPLANATION:

S=v t ∵t=T

T=S/v

S=2πR

T=2πR/v

Discussion

48.	A parallel plate air capacitor has a capacitance C. If air between the plates isexhausted completely, the new capacitancewould become(i) zero(ii) C(iii) slightly more than C( slightly less than C
Answer» Answer: less than c because CAPACITANCE= (K A)/D HERE K IS THE DIELCTRIC CONSTANT OF THE MEDIUM BETWEEN PLATES IF AIR IS REMOVED , DIECTRIC CONSTANT OF VACUM IS LESS THAN AIR THUS CAPCITANCE IS REDUCED

48.

A parallel plate air capacitor has a capacitance C. If air between the plates isexhausted completely, the new capacitancewould become(i) zero(ii) C(iii) slightly more than C( slightly less than C

Answer»

Answer:

less than c because

CAPACITANCE= (K A)/D

HERE K IS THE DIELCTRIC CONSTANT OF THE MEDIUM BETWEEN PLATES

IF AIR IS REMOVED , DIECTRIC CONSTANT OF VACUM IS LESS THAN AIR THUS CAPCITANCE IS REDUCED

Discussion

49.	an air conditioner of 2kw is operated for 8 hrs daily . what will be the energy consumed for a day and also for the entire may month?
Answer» Answer: 16 and 480 Explanation: ENERGY CONSUMED=POWER×TIME : =2×8 =16 J ENERGY CONSUMED ENTIRE MONTH 16×30=480J

49.

an air conditioner of 2kw is operated for 8 hrs daily . what will be the energy consumed for a day and also for the entire may month?

Answer»

Answer:

16 and 480

Explanation:

ENERGY CONSUMED=POWER×TIME

: =2×8

=16 J

ENERGY CONSUMED ENTIRE MONTH

16×30=480J

Discussion

50.	Solve karo saalo dekhy kis mai hai kitna dam
Answer» ANSWER: No, because you are bad BOY and use ABUSIVE language

50.

Solve karo saalo dekhy kis mai hai kitna dam

Answer»

ANSWER:

No, because you are bad BOY and use ABUSIVE language

Discussion

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