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Class 12th Physics Nuclei

The activity of a radioactive material is 2.56 * 10^-³ Ci. If the half life of the material is 5 days, after how many days the activity will become 2 * 10^-⁵ Ci?

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alok kumar singh

Contributor-Level 10

A₀ = 2.56 * 10^-3C_i

A = 2 * 10^-15C_i

A = ^{$\frac{A_{0}}{2^{n}}$}

$\Rightarrow 2 * 1 0^{- 5} = \frac{2.56 * 1 0^{- 3}}{2^{n}}$

2ⁿ = 128

n = 7

&nbs

...more

A₀ = 2.56 * 10^-3C_i

A = 2 * 10^-15C_i

A = ^{$\frac{A_{0}}{2^{n}}$}

$\Rightarrow 2 * 1 0^{- 5} = \frac{2.56 * 1 0^{- 3}}{2^{n}}$

2ⁿ = 128

n = 7

t = 7 * t_1/2

= 7 * 5 = 35 days

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11 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Five

A body of mass M at rest explodes into pieces, in the ratio of masses 1 : 1: 2. Two smaller pieces fly off perpendicular to each other with velocities of 30 ms^-¹ and 40 ms-¹ respectively. The velocity of the third piece will be:

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alok kumar singh

Contributor-Level 10

Let 'v' be the velocity of third piece

com along x-axis

M * 0 = M₁v_x + M₃v'_x

$\Rightarrow 0 = M_{1} * 30 + M_{3} v' x$

$v_{x} 1 = \frac{- M_{1}}{M_{3}} * 30$

$\frac{- 1}{2} * 30$

$v_{x}^{1} = - 15 m / s e c$

Com along y-axis

&nb

...more

Let 'v' be the velocity of third piece

com along x-axis

M * 0 = M₁v_x + M₃v'_x

$\Rightarrow 0 = M_{1} * 30 + M_{3} v' x$

$v_{x} 1 = \frac{- M_{1}}{M_{3}} * 30$

$\frac{- 1}{2} * 30$

$v_{x}^{1} = - 15 m / s e c$

Com along y-axis

M * 0 = M₂v_y + M₃v'_y

v_y = $\frac{- M_{2}}{M_{3}} v_{y}$

v_y = $\frac{- 1}{2} * 40$

v_y = -20m/sec

^{$\vec{v} = - 15 \hat{i} - 20 \hat{j}$}

v = $\sqrt[]{1 5^{2} + 2 0^{2}} = \sqrt[]{225 + 400} = 25 m / s e c$

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11 months ago

Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Thirteen

Consider the reaction and predict the major product.

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

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11 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Three

Two balls A and B placed at the top of 180m tall tower. Ball A is from the top at t = 0s. Ball B is thrown vertically down with an initial velocity 'u' at t = 2s. After a certain time, both balls meet 100m above the ground. Find the value of 'u' in ms^-¹.

[use g = 10 ms^-²] :

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alok kumar singh

Contributor-Level 10

Let 't' be the time taken by B to meat A.

for A, u = 0

t = [2 + t]

h = 80m

a = 10 m/sec²

h = ut + $\frac{1}{2} a t^{2}$

80 = 0 * t + $\frac{1}{2} (10) {(2 + t)}^{2}$

16 = (2 + t)²

t + 2 = 4

t = 2sec

Now for B

h = ut + $\frac{1}{2} a t^{2}$

...more

Let 't' be the time taken by B to meat A.

for A, u = 0

t = [2 + t]

h = 80m

a = 10 m/sec²

h = ut + $\frac{1}{2} a t^{2}$

80 = 0 * t + $\frac{1}{2} (10) {(2 + t)}^{2}$

16 = (2 + t)²

t + 2 = 4

t = 2sec

Now for B

h = ut + $\frac{1}{2} a t^{2}$

$? 80 = u * 2 + \frac{1}{2} * 10 t^{2}$

80 = 2u + 5 * 2 * 2

80 = 2u + 20

2u = 60

u = 30m/sec

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New answer posted

11 months ago

Class 12th System of Particles and Rotational Motion

A spherical of 1 kg mass and radius R is rolling with angular speed $ω$ on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin O is $\frac{a}{3} R^{2} ω .$ The value of a will be

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Payal Gupta

Contributor-Level 10

L₀ = RMV_cm + I_cm $ω$

= RM $(R ω) + \frac{2}{3} M R^{2} ω$

L₀ = $\frac{5}{3} M R^{2} ω$

= $\frac{5}{3} * 1 R^{2} ω = \frac{5}{3} R^{2} ω = \frac{a}{3} R^{2} ω$

a = 5

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New answer posted

11 months ago

Class 12th Nuclei

The activity of a radioactive material is 2.56 * 10^-3 Ci. If the half life of the material is 5 days, after how many days the activity will become 2 * 10^-5 Ci?

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Payal Gupta

Contributor-Level 10

A₀ = 2.56 * 10^-3C_i

A = 2 * 10^-15C_i

A = $\frac{A_{0}}{2^{n}}$

$\Rightarrow 2 * 1 0^{- 5} = \frac{2.56 * 1 0^{- 3}}{2^{n}}$

2ⁿ = 128

n = 7

t = 7 * t_1/2

= 7 * 5 = 35 days

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New answer posted

11 months ago

Class 12th Laws of Motion

A body of mass M at rest explodes into pieces, in the ratio of masses 1 : 1: 2. Two smaller pieces fly off perpendicular to each other with velocities of 30 ms^-1 and 40 ms^-1 respectively. The velocity of the third piece will be

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

Let 'v' be the velocity of third piece

com along x-axis

M * 0 = M₁v_x + M₃v'_x

$\Rightarrow 0 = M_{1} * 30 + M_{3} v' x$

$v_{x} 1 = \frac{- M_{1}}{M_{3}} * 30$

= $\frac{- 1}{2} * 30$

$v_{x}^{1} = - 15 m / s e c$

Com along y-axis

M * 0 = M₂vy + M₃v'y

v_y = $\frac{- M_{2}}{M_{3}} v_{y}$

v_y = $\frac{- 1}{2} * 40$

v_y = 20m/sec

$\vec{v} = - 15 \hat{i} - 20 \hat{j}$

v = $\sqrt[]{1 5^{2} + 2 0^{2}} = \sqrt[]{225 + 400} = 25 m / s e c$

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New question posted

11 months ago

Class 12th Laws of Motion

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New answer posted

11 months ago

Class 12th Motion in a Straight Line

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Payal Gupta

Contributor-Level 10

Let 't' be the time taken by B to meat

for A, u = 0

t = [2 + t]

h = 80m

a = 10 m/sec²

h = ut + $\frac{1}{2} a t^{2}$

80 = 0 * t + $\frac{1}{2} (10) {(2 + t)}^{2}$

16 = (2 + t)²

t + 2 = 4

t = 2sec

Now for B

h = ut + $\frac{1}{2} a t^{2}$

$\Rightarrow 80 = u * 2 + \frac{1}{2} * 10 t^{2}$

80 = 2u + 5 * 2 * 2

80 = 2u + 20

2u = 60

u = 30m/sec

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New answer posted

11 months ago

Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Thirteen

The correct sequential order fo the reagents for the given reaction is

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

0 0

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