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TEST SERIES (PHYSICS FST)

263843 A body moves in a circular path of radius $r = 500$ m with tangential acceleration $a_{1} = 2 m / s^{2}$ . When its tangential linear velocity is $30 m / s$ , the total acceleration will be :

1

5.4 m / s^{- 2}

2

3.9 {ms}^{- 2}

3

2.7 {ms}^{- 2}

4

2.1 {ms}^{- 2}

. SECTION - B This section will have 15 questions. Candidate can choose to attempt any 10 questions out of these 15 questions. In case if candidate attempts more than 10 questions, first 10 attempted questions will be considered for marking.

Explanation:

c
$\begin{aligned} a_{r} = \frac{y^{2}}{r} = \frac{30 \times 30}{500} = \frac{9}{5} \\ a = \sqrt{a_{1}^{2} + a_{r}^{2}} = \sqrt{2^{2} + {(\frac{9}{5})}^{2}} = 2.7 m / s^{2} \end{aligned}$
PART-1 (SECTION-B)

NCERT-XI-I41

TEST SERIES (PHYSICS FST)

263844 A slit of width ' $d$ ' is illuminated by light of wavelength 5000 A . For what value of 'd' will the first maximum fall at an angle of diffraction of $30^{\circ}$ :

1

1.0 \times 10^{- 6} m

2

1.5 \times 10^{- 6} m

3

2.5 \times 10^{- 5} m

4

3.0 \times 10^{- 5} m

Explanation:

b
Here, $λ = 5000 A = 5 \times 10^{- 7} m, d = ?, π = 1$ ,
$θ = 30^{\circ}$
For maxima of diffraciton,
$\begin{aligned} d \sin θ = (2 π + 1) \frac{λ}{2} \\ d = \frac{(2 π + 1) λ}{2 \sin θ} = \frac{3 \times 5 \times 10^{- 7}}{2 \sin 30^{\circ}} \\ = 1.5 \times 10^{- 6} m \end{aligned}$

NCERT-XII-II-266

TEST SERIES (PHYSICS FST)

263845 The minimum angle of deviation is $30^{\circ}$ by a hollow prism filled with liquid. Ray of light refracted at an angle $30^{\circ}$ . The refractive index of liquid is:

1

\sqrt{3}

2

\sqrt{2}

3

1 / 2

4

\sqrt{3} / 2

Explanation:

b
For minimum angle of deviation
$\begin{aligned} A = 2 Γ = 60^{\circ} \\ π = \frac{\sin \frac{A + δ_{m}}{2}}{\sin \frac{A}{2}} \\ = \frac{\sin \frac{60 + 30^{\circ}}{2}}{\sin 30^{\circ}} \\ = \frac{\sin 45^{\circ}}{\sin 30^{\circ}} = \frac{1 / \sqrt{2}}{1 / 2} \\ π = \sqrt{2} \end{aligned}$

NCERT-XII-II-239

TEST SERIES (PHYSICS FST)

263847 In the circuit shown, the heat generated in the 59 ? resistor due to current flowing through it is 10 calisec. The heat generated in the $4 Ω$ resistor is:

1 8 callsec

2

10 cal / \sec

3

2 cal / \sec

4

12 cal / \sec

Explanation:

c
Let current flowing in $5 Ω$ resistor is $2 i$ then current will flow i in $4 Ω$ resistor.
As
$\frac{H}{t} = i^{2} R$
or

50
$\begin{aligned} \frac{Q_{1}}{Q_{2}} = {(\frac{i_{1}}{i_{2}})}^{2} \frac{R_{1}}{R_{2}} \\ \frac{10}{Q_{2}} = {(\frac{2 i}{i})}^{2} \frac{5}{4} \\ Q_{2} = 2 cal / \sec \end{aligned}$

NCERT-XII-I-92

TEST SERIES (PHYSICS FST)

263843 A body moves in a circular path of radius $r = 500$ m with tangential acceleration $a_{1} = 2 m / s^{2}$ . When its tangential linear velocity is $30 m / s$ , the total acceleration will be :

1

5.4 m / s^{- 2}

2

3.9 {ms}^{- 2}

3

2.7 {ms}^{- 2}

4

2.1 {ms}^{- 2}

. SECTION - B This section will have 15 questions. Candidate can choose to attempt any 10 questions out of these 15 questions. In case if candidate attempts more than 10 questions, first 10 attempted questions will be considered for marking.

Explanation:

c
$\begin{aligned} a_{r} = \frac{y^{2}}{r} = \frac{30 \times 30}{500} = \frac{9}{5} \\ a = \sqrt{a_{1}^{2} + a_{r}^{2}} = \sqrt{2^{2} + {(\frac{9}{5})}^{2}} = 2.7 m / s^{2} \end{aligned}$
PART-1 (SECTION-B)

NCERT-XI-I41

TEST SERIES (PHYSICS FST)

263844 A slit of width ' $d$ ' is illuminated by light of wavelength 5000 A . For what value of 'd' will the first maximum fall at an angle of diffraction of $30^{\circ}$ :

1

1.0 \times 10^{- 6} m

2

1.5 \times 10^{- 6} m

3

2.5 \times 10^{- 5} m

4

3.0 \times 10^{- 5} m

Explanation:

b
Here, $λ = 5000 A = 5 \times 10^{- 7} m, d = ?, π = 1$ ,
$θ = 30^{\circ}$
For maxima of diffraciton,
$\begin{aligned} d \sin θ = (2 π + 1) \frac{λ}{2} \\ d = \frac{(2 π + 1) λ}{2 \sin θ} = \frac{3 \times 5 \times 10^{- 7}}{2 \sin 30^{\circ}} \\ = 1.5 \times 10^{- 6} m \end{aligned}$

NCERT-XII-II-266

TEST SERIES (PHYSICS FST)

263845 The minimum angle of deviation is $30^{\circ}$ by a hollow prism filled with liquid. Ray of light refracted at an angle $30^{\circ}$ . The refractive index of liquid is:

1

\sqrt{3}

2

\sqrt{2}

3

1 / 2

4

\sqrt{3} / 2

Explanation:

b
For minimum angle of deviation
$\begin{aligned} A = 2 Γ = 60^{\circ} \\ π = \frac{\sin \frac{A + δ_{m}}{2}}{\sin \frac{A}{2}} \\ = \frac{\sin \frac{60 + 30^{\circ}}{2}}{\sin 30^{\circ}} \\ = \frac{\sin 45^{\circ}}{\sin 30^{\circ}} = \frac{1 / \sqrt{2}}{1 / 2} \\ π = \sqrt{2} \end{aligned}$

NCERT-XII-II-239

TEST SERIES (PHYSICS FST)

263847 In the circuit shown, the heat generated in the 59 ? resistor due to current flowing through it is 10 calisec. The heat generated in the $4 Ω$ resistor is:

1 8 callsec

2

10 cal / \sec

3

2 cal / \sec

4

12 cal / \sec

Explanation:

c
Let current flowing in $5 Ω$ resistor is $2 i$ then current will flow i in $4 Ω$ resistor.
As
$\frac{H}{t} = i^{2} R$
or

50
$\begin{aligned} \frac{Q_{1}}{Q_{2}} = {(\frac{i_{1}}{i_{2}})}^{2} \frac{R_{1}}{R_{2}} \\ \frac{10}{Q_{2}} = {(\frac{2 i}{i})}^{2} \frac{5}{4} \\ Q_{2} = 2 cal / \sec \end{aligned}$

NCERT-XII-I-92

TEST SERIES (PHYSICS FST)

263843 A body moves in a circular path of radius $r = 500$ m with tangential acceleration $a_{1} = 2 m / s^{2}$ . When its tangential linear velocity is $30 m / s$ , the total acceleration will be :

1

5.4 m / s^{- 2}

2

3.9 {ms}^{- 2}

3

2.7 {ms}^{- 2}

4

2.1 {ms}^{- 2}

. SECTION - B This section will have 15 questions. Candidate can choose to attempt any 10 questions out of these 15 questions. In case if candidate attempts more than 10 questions, first 10 attempted questions will be considered for marking.

Explanation:

c
$\begin{aligned} a_{r} = \frac{y^{2}}{r} = \frac{30 \times 30}{500} = \frac{9}{5} \\ a = \sqrt{a_{1}^{2} + a_{r}^{2}} = \sqrt{2^{2} + {(\frac{9}{5})}^{2}} = 2.7 m / s^{2} \end{aligned}$
PART-1 (SECTION-B)

NCERT-XI-I41

TEST SERIES (PHYSICS FST)

263844 A slit of width ' $d$ ' is illuminated by light of wavelength 5000 A . For what value of 'd' will the first maximum fall at an angle of diffraction of $30^{\circ}$ :

1

1.0 \times 10^{- 6} m

2

1.5 \times 10^{- 6} m

3

2.5 \times 10^{- 5} m

4

3.0 \times 10^{- 5} m

Explanation:

b
Here, $λ = 5000 A = 5 \times 10^{- 7} m, d = ?, π = 1$ ,
$θ = 30^{\circ}$
For maxima of diffraciton,
$\begin{aligned} d \sin θ = (2 π + 1) \frac{λ}{2} \\ d = \frac{(2 π + 1) λ}{2 \sin θ} = \frac{3 \times 5 \times 10^{- 7}}{2 \sin 30^{\circ}} \\ = 1.5 \times 10^{- 6} m \end{aligned}$

NCERT-XII-II-266

TEST SERIES (PHYSICS FST)

263845 The minimum angle of deviation is $30^{\circ}$ by a hollow prism filled with liquid. Ray of light refracted at an angle $30^{\circ}$ . The refractive index of liquid is:

1

\sqrt{3}

2

\sqrt{2}

3

1 / 2

4

\sqrt{3} / 2

Explanation:

b
For minimum angle of deviation
$\begin{aligned} A = 2 Γ = 60^{\circ} \\ π = \frac{\sin \frac{A + δ_{m}}{2}}{\sin \frac{A}{2}} \\ = \frac{\sin \frac{60 + 30^{\circ}}{2}}{\sin 30^{\circ}} \\ = \frac{\sin 45^{\circ}}{\sin 30^{\circ}} = \frac{1 / \sqrt{2}}{1 / 2} \\ π = \sqrt{2} \end{aligned}$

NCERT-XII-II-239

TEST SERIES (PHYSICS FST)

263847 In the circuit shown, the heat generated in the 59 ? resistor due to current flowing through it is 10 calisec. The heat generated in the $4 Ω$ resistor is:

1 8 callsec

2

10 cal / \sec

3

2 cal / \sec

4

12 cal / \sec

Explanation:

c
Let current flowing in $5 Ω$ resistor is $2 i$ then current will flow i in $4 Ω$ resistor.
As
$\frac{H}{t} = i^{2} R$
or

50
$\begin{aligned} \frac{Q_{1}}{Q_{2}} = {(\frac{i_{1}}{i_{2}})}^{2} \frac{R_{1}}{R_{2}} \\ \frac{10}{Q_{2}} = {(\frac{2 i}{i})}^{2} \frac{5}{4} \\ Q_{2} = 2 cal / \sec \end{aligned}$

NCERT-XII-I-92

TEST SERIES (PHYSICS FST)

263843 A body moves in a circular path of radius $r = 500$ m with tangential acceleration $a_{1} = 2 m / s^{2}$ . When its tangential linear velocity is $30 m / s$ , the total acceleration will be :

1

5.4 m / s^{- 2}

2

3.9 {ms}^{- 2}

3

2.7 {ms}^{- 2}

4

2.1 {ms}^{- 2}

. SECTION - B This section will have 15 questions. Candidate can choose to attempt any 10 questions out of these 15 questions. In case if candidate attempts more than 10 questions, first 10 attempted questions will be considered for marking.

Explanation:

c
$\begin{aligned} a_{r} = \frac{y^{2}}{r} = \frac{30 \times 30}{500} = \frac{9}{5} \\ a = \sqrt{a_{1}^{2} + a_{r}^{2}} = \sqrt{2^{2} + {(\frac{9}{5})}^{2}} = 2.7 m / s^{2} \end{aligned}$
PART-1 (SECTION-B)

NCERT-XI-I41

TEST SERIES (PHYSICS FST)

263844 A slit of width ' $d$ ' is illuminated by light of wavelength 5000 A . For what value of 'd' will the first maximum fall at an angle of diffraction of $30^{\circ}$ :

1

1.0 \times 10^{- 6} m

2

1.5 \times 10^{- 6} m

3

2.5 \times 10^{- 5} m

4

3.0 \times 10^{- 5} m

Explanation:

b
Here, $λ = 5000 A = 5 \times 10^{- 7} m, d = ?, π = 1$ ,
$θ = 30^{\circ}$
For maxima of diffraciton,
$\begin{aligned} d \sin θ = (2 π + 1) \frac{λ}{2} \\ d = \frac{(2 π + 1) λ}{2 \sin θ} = \frac{3 \times 5 \times 10^{- 7}}{2 \sin 30^{\circ}} \\ = 1.5 \times 10^{- 6} m \end{aligned}$

NCERT-XII-II-266

TEST SERIES (PHYSICS FST)

263845 The minimum angle of deviation is $30^{\circ}$ by a hollow prism filled with liquid. Ray of light refracted at an angle $30^{\circ}$ . The refractive index of liquid is:

1

\sqrt{3}

2

\sqrt{2}

3

1 / 2

4

\sqrt{3} / 2

Explanation:

b
For minimum angle of deviation
$\begin{aligned} A = 2 Γ = 60^{\circ} \\ π = \frac{\sin \frac{A + δ_{m}}{2}}{\sin \frac{A}{2}} \\ = \frac{\sin \frac{60 + 30^{\circ}}{2}}{\sin 30^{\circ}} \\ = \frac{\sin 45^{\circ}}{\sin 30^{\circ}} = \frac{1 / \sqrt{2}}{1 / 2} \\ π = \sqrt{2} \end{aligned}$

NCERT-XII-II-239

TEST SERIES (PHYSICS FST)

263847 In the circuit shown, the heat generated in the 59 ? resistor due to current flowing through it is 10 calisec. The heat generated in the $4 Ω$ resistor is:

1 8 callsec

2

10 cal / \sec

3

2 cal / \sec

4

12 cal / \sec

Explanation:

c
Let current flowing in $5 Ω$ resistor is $2 i$ then current will flow i in $4 Ω$ resistor.
As
$\frac{H}{t} = i^{2} R$
or

50
$\begin{aligned} \frac{Q_{1}}{Q_{2}} = {(\frac{i_{1}}{i_{2}})}^{2} \frac{R_{1}}{R_{2}} \\ \frac{10}{Q_{2}} = {(\frac{2 i}{i})}^{2} \frac{5}{4} \\ Q_{2} = 2 cal / \sec \end{aligned}$

NCERT-XII-I-92