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10 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Six

Two identical steel cubes (masses 50g, side 1cm) collide head-on face to face with a speed of 10cm/s each. Find the maximum compression of each. Young's modulus for steel = Y= 2 * 10¹¹ N/m².

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

m =50g = 50 $* 10^{3} k g$

Side = 1cm = 0.01m

Speed v = 0.1m/s

Young's modulus= 2 $* 10^{11} N / m$ ²

According to the formula

F/A= Y $\frac{? L}{L}$

And F= K $? L$ where K is the compression in the spring.

K= YA/L = YL

Initial KE= 2 (1/2mv²)= 5 $* 10^{- 4} J$

Final PE= 2 (1/2)K ( $? L$ )²

$? L = \sqrt[]{\frac{K E}{E}} = \sqrt[]{\frac{K E}{Y L}}$ = $\sqrt[]{\frac{5 * 10^{- 4}}{2 * 10^{11} * 0.1}} = 1.58 * 10^{- 7} m$

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10 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Six

A rocket accelerates straight up by ejecting gas downwards. In a small time interval ?t, it ejects a gas of mass ?m at a relative speed u. Calculate KE of the entire system at t + ?t and t and show that the device that ejects gas does work = (1/2) ?mu² in this time interval (neglect gravity).

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Let M be the mass of the rocket at any time t and v₁ the velocity of the rocket at the same time t

Let? m = mass of gas ejected in? t time

Relative speed of the gas ejected =u

KE +? t = KE of rocket +KE of gas

= ½ (M-? m) (v+? v)² + ½? m (v-u)²

KE_t= KE of rocket at time t= ½ Mv²

So? K = KE_t+? t -KE_t

= (M? v=? mu)v+1/2? mu²

Since action and reaction forces are equal

M? v/? t=? m/? t|u|

M? v=? m u

So? K= ½? mu²

? K=? W

? W=1/2? mu²

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10 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Six

A curved surface is shown in Fig. The portion BCD is free of friction. There are three spherical balls of identical radii and masses. Balls are released from rest one by one from A which is at a slightly greater height than C With the surface AB, ball 1 has large enough friction to cause rolling down without slipping; ball 2 has a small friction and ball 3 has a negligible friction.

(a) For which balls is total mechanical energy conserved?

(b) Which ball(s) can reach D?

(c) For balls which do not reach D, which of the balls can reach back A?

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

(a) As ball 1 is rolling down without slipping there is no dissipation of energy hence, total mechanical energy is conserved. Ball 3 is having negligible friction hence, there is no loss of energy.

(b) Ball 1 acquires rotational energy, ball 2 loses energy by friction. They cannot cross at C. Ball 3 can cross over.

(c) Ball 1, 2 turn back before reaching C. Because of loss of energy, ball 2 cannot reach back to A. Ball 1 has a rotational motion in “wrong” sense when it reaches B. It cannot roll back to A, because of kinetic friction.

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10 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Six

A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 30° by a force of 10 N parallel to the inclined surface in fig

The coefficient of friction between block and the incline is 0.1. If the block is pushed up by 10 m along the incline, calculate

(a) work done against gravity

(b) work done against force of friction

(c) increase in potential energy

(d) increase in kinetic energy

(e) work done by applied force.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

(a) work done= increase in PE

= mg (vertical distance travelled)

= mg (s)sin $θ = 1 * 10 * 10 s i n 30$ = 50J

(b) work done against friction = fs

= $μ N (s) = μ m g c o s θ$

= 0.1 $* 1 * 10 * c o s 30 * 10 = 8.66 J$

(c) increase in PE =mgh

=1 $* 10 * 10 * s i n 30$

$= 100 (\frac{1}{2}) = 50 J$

(d) according to work energy theorem W= change in KE

= -mgh-fs+FS

= -50-8.66+10 (10)

= 41.34J

(e) force f = FS

= 10 (10)= 100J

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10 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter Three

Electronic configuration of some elements is given in Column I and their electron gain enthalpies are given in Column II. Match the electronic configuration with electron gain enthalpy.

Column (I)	Column (II)
Electronic configuration	Electron gain enthalpy/kJ mol^–1
(i) 1s²2s²sp⁶	(A) – 53
(ii) 1s² 2s² 2p⁶ 3s¹	(B) – 328
(iii) 1s²2s² 2p⁵	(C) – 141
(iv) 1s² 2s² 2p⁴	(D) + 48

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alok kumar singh

Contributor-Level 10

This is a Matching Type Questions as classified in NCERT Exemplar

i d + 48

ii a &n

...more

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10 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter Three

Electronic configuration of some elements is given in Column I and their electron gain enthalpies are given in Column II. Match the electronic configuration with electron gain enthalpy.

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New answer posted

10 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter Three

Match the correct ionisation enthalpies and electron gain enthalpies of the following elements.

Elements		?H₁	? H₂	?_egH
(i) Most reactive non metal	A	419	3051	– 48
(ii) Most reactive metal	B	i b	i b	i b
(iii) Least reactive element	C	738	1451	– 40
(iv) Metal forming binary halide	D	2372	5251	+ 48

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alok kumar singh

Contributor-Level 10

This is a Matching Type Questions as classified in NCERT Exemplar

i ) b 1681 3374 – 328

ii ) a 419 3051 – 48

iii)

...more

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10 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter Three

Match the correct atomic radius with the element.

Element	Atomic Radius (pm)
Be	74
C	88
O	111
B	77
N	66

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alok kumar singh

Contributor-Level 10

This is a Matching Type Questions as classified in NCERT Exemplar

Be 111

...more

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10 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter Three

Assertion (A) : Electron gain enthalpy becomes less negative as we go down a group.

Reason (R) : Size of the atom increases on going down the group and the added electron would be farther from the nucleus.

(i) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

(ii) Assertion and reason both are correct statements and reason is correct explanation for assertion.

(iii) Assertion and reason both are wrong statements.

(iv) Assertion is wrong statement but reason is correct statement.

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alok kumar singh

Contributor-Level 10

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Option (ii)

The electron gain enthalpy decreases down the group due to the increase in the atomic size which leads to decrease in the effective nuclear charge.

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10 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter Three

Assertion (A) : Boron has a smaller first ionisation enthalpy than beryllium.

Reason (R) : The penetration of a 2s electron to the nucleus is more than the 2p electron hence 2p electron is more shielded by the inner core of electrons than the 2s electrons.

(i) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

(ii) Assertion is correct statement but reason is wrong statement.

(iii) Assertion and reason both are correct statements and reason is correct explanation for assertion.

(iv) Assertion and reason both are wrong statements

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alok kumar singh

Contributor-Level 10

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Option (iii)

The Be has the electronic configuration of [He]2s² while B has the electronic configuration of [He]2s²2p¹ . As in Be the outermost orbital i.e. 2s is fully filled thus it provides extra stability to the Be compared to B. Apart from this 2s electron is more penetrated compared to the 2p electron for which 2p electron faces more shielding effect than that of 2s electron.

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