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This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b)

Explanation- the moderator used have light nuclei . because when we use heavy nuclei then there will be collision of proton which does not slow down the process.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b)

Explanation – a nucleus contain more number of neutron than protons because the repulsive force is more if it contain more proton and become more unstable.

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

Inductive reactance XL= 2 $\pi \vartheta$ L

And capacitive reactance Xc= $\frac{1}{2\pi \vartheta \mathrm{C}}$

For very high frequencies XL= infinity and Xc=0

When reactance of the circuit is infinite it will be considered as open circuit . when reactance of a circuit is zero it will be considered as short circuited.

So C1, C2 = shorted and L1, L2=opened

So R_eff= R₁+R₂

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation – tritium contain 1 proton and 2 neutron but when it disintegrates and remove 1 beta particle it will now contain 2 proton and 1 neutron so it will change into ₂He³ but its energy will be less from it.

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

If we consider a L-C circuit analogous to a harmonically oscillating spring block system. The electrostatic energy $\frac{1}{2}$ CV2 is analogous to potential energy and energy associated with moving charges (current) that is magnetic energy $\frac{1}{2}$ LI2 is analogous to kinetic energy.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation-All points with the same optical path length must have the same phase.

So – $\sqrt[]{{\mu }_r{\epsilon }_r}AE$   =BC-–  $\sqrt[]{{\mu }_r{\epsilon }_r}CD$

BC= $\sqrt[]{{\mu }_r{\epsilon }_r}$ (CD-AE)

BC>0, si must be greater than AD

But in other figure

–  $\sqrt[]{{\mu }_r{\epsilon }_r}AE=BC–\mathrm{ }\mathrm{ }\sqrt[]{{\mu }_r{\epsilon }_r}CD$

So BC= –  $\sqrt[]{{\mu }_r{\epsilon }_r}\left(CD-AE\right)$

But clearly here BE is less than zero

To proving snells law we know that

BC=ACsin $\theta$ and CD-AE=ACsin $\theta$

So n= sini/sinr

The eligibility criteria for the certification programme depends upon the specialisation that the candidate wishes to opt for. The eligibility criteria for various specialisations can require candidates to either have cleared Class 10 or Class 12. Candidates can learn more about the eligibility and the selection criteria for various specialisations by visiting the official website and checking the long term or short term course information. 

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer –a 

Explanation- $\beta$ ⁺ decay represents

_zX^{A $\to$} _z-1Y^A+₁e^o+v+Q₂

Q₂= [m_n (_zX^A)-m_n (_z-1Y^A)-m_e]c²

= (M_x-M_y-2m_e)c²

^{$\beta$ -} decay represents

_zX^{A $\to$} _z+1Y^A+-₁e^o+v+ $\propto$ ₁

_{$\propto$ 1}= [m_n (_zX^A)-m_n (_z+1Y^A)-m_e]c²

= (M_x-M_y)c²

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

A fresh graduate after studying BSc Food Science and Nutrition can get an average salary of INR 2.5 to 6 LPA in entry level positions, but with experience and additional qualifications, you can easily earn up to INR 10 LPA. 

Recommended Link: BSc Career Scope and Salary in India

No, on the digital SAT, everyone gets a different question. Based on answers to questions, the difficulty or ease of the upcoming questions is adjusted. Thus, the question tested for one test taker will be different for the other test taker. Similar or patterned questions may be asked but not the same content questions.  

Everyone got the same questions on SAT exam when the exam was a pencil-paper based test. 

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b)

Explanation – because alpha and beta particle have some charge and mass so their energy level will change but in case of gamma particle it does not contain any charge.

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

Students need not worry even if you do not have enough time left for MBSE 12th board exam preparation. Check the tips given below for MBSE HSSLC board exam preparation.

• Learn all important topics
• Make a proper study plan
• Take help from teachers or friends if needed
• Practice past year question papers
• Manage Time 
• Stay calm, healthy and positive

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- as the refractive index of the class $\mu$ , the path difference will be calculated as $?x$ =2dsin $\theta$ +( $\mu -1$ )L

For principal maxima ,(path difference is zero)

2dsin $\theta$ ₀+( $\mu -1$ )L=0

Sin $\theta$ ₀= - $\frac{L(\mu -1)}{2d}$ = $\frac{-L\left(0.5\right)}{2d}$

Sin $\theta$ ₀=-1/16

OP=Dtan $\theta$ ₀= Dsin $\theta$ ₀=-D/16

For pat $?$ h difference $?\frac{\lambda }{2}$

2dsin $\theta$ ₁+0.5L= $?\frac{\lambda }{2}$

Sin $\theta$ ₁= $\frac{?\frac{\lambda }{2}-0.5L}{2d}$ = $\frac{?\frac{\lambda }{2}-\frac{d}{8}}{2d}$

= $\frac{\frac{\lambda }{2}-\frac{\lambda }{8}}{2\lambda }$ = $?$ 1/4 -1/16

So two possible values $\frac{1}{4}-\frac{1}{16}=\frac{3}{16}$  and- $\frac{1}{4}-\frac{1}{16}$ = $-\frac{5}{16}$

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