Explore topic-wise InterviewSolutions in .

(i) Is Mr. Rahim at home 

(ii) It’s Ram. May I see you for five minutes 

(iii) It’s an important personal matter 

(iv) I have an urgent appointment with my boss 

(v) See you tomorrow at 9 a.m

The valency of N is 3

Fatty acyl synthase has three important characteristics : It is essential, but not rate-limiting, for fatty acid synthesis. It is not subject to short term control. ACP and the catalytic activities are on a single contiguous protein (257 kDa).

ᴀ ᴛᴇɴᴅᴏɴ ɪs ᴀ ᴛᴏᴜɢʜ, ғʟᴇxɪʙʟᴇ ʙᴀɴᴅ ᴍᴀᴅᴇ ᴏғ ғɪʙʀᴏᴜs ᴄᴏɴɴᴇᴄᴛɪᴠᴇ ᴛɪssᴜᴇ, ᴀɴᴅ ғᴜɴᴄᴛɪᴏɴs ᴛᴏ ᴄᴏɴɴᴇᴄᴛ ᴍᴜsᴄʟᴇ ᴛᴏ ʙᴏɴᴇ. ᴊᴏɪɴᴛs ᴀʀᴇ ᴛʜᴇ ʙᴏɴᴇ ᴀʀᴛɪᴄᴜʟᴀᴛɪᴏɴs ᴀʟʟᴏᴡɪɴɢ ᴍᴏᴠᴇᴍᴇɴᴛ. ᴀ ʟɪɢᴀᴍᴇɴᴛ ɪs ᴀ ᴅᴇɴsᴇ, ᴡʜɪᴛᴇ ʙᴀɴᴅ ᴏғ ғɪʙʀᴏᴜs ᴇʟᴀsᴛɪᴄ ᴛɪssᴜᴇ. ʟɪɢᴀᴍᴇɴᴛs ᴄᴏɴɴᴇᴄᴛ ᴛʜᴇ ᴇɴᴅs ᴏғ ʙᴏɴᴇs ᴛᴏɢᴇᴛʜᴇʀ ɪɴ ᴏʀᴅᴇʀ ᴛᴏ ғᴏʀᴍ ᴀ ᴊᴏɪɴᴛ.

(i) May I help you 

(ii) Is something wrong with this book 

(iii) When did you buy it 

(iv) Do you have the biu 

(v) and the book. Kindly exchange it

budding of reproduction bud formation take place on parent body.

Release of CO₂ is not photosynthesis process. 

Uro II is related to air pollution.

Right food chain is grass → deer → lion

Respiration function is necessary for all organism.

Asian Regional Forum is related with  ASEAN

Bangaldesh  was Afro-Asian countries conference held in 1955.

Soviet Union disintegrated in 25 December 1991

- Blood passes through filtration units in the kidney called nephron 

- Passes through glomerulus in the Bowman's capsule - Ultra filtration

- Filtrate initially has glucose, amino acids, water, salts and nitrogenous waste 

- Reabsorption – Water (as per the need of the body), Glucose and amino acids are all reabsorbed – 

- Secretion of excess water, salts and urea (nitrogenous waste) which makes up the urine

(i) D, As it is on the left side of the table in group 2 

(ii) C, as it is in the group 18/ Noble gas 

(iii) E, as we move from left to right across a period, atomic radius decreases

a. Heating of lead nitrate; and electrolysis of acidified water

b. 2AgCl_(s) \(\overset{Sunlight}{\longrightarrow}\) 2Ag_(s) + Cl_2 (g) 

(No deduction for not mentioning state of reactants and products.)

Heating of lead nitrate; and electrolysis of acidified water

Cortisol can help control blood sugar levels, regulate metabolism, help reduce inflammation, and assist with memory formulation. It has a controlling effect on salt and water balance and helps control blood pressure.

(D) 1

Overall order of reaction can be decided by the data given t_75% = 2t_50% 

∴ It is a first order reaction with respect to P. 

From graph [Q] is linearly decreasing with time, i.e. order of reaction with respect to Q is zero and the rate expression is r = k [P]¹ [Q]⁰ .

A = {1,2,3,.....,100}

R = {(a,b) ∈ A x A | b = a³}

∴ R = {(1,1),(2,8),(3,27),(4,64)}

∴ Domain of R is D = {1,2,3,4} &

Range of R is R_R = {1,8,27,64}

(a) 6 

(b) 4

x = {0, 1, 2, 3, 4, 5} 

y = x + 3 

When x = 0 ⇒ y = 0 + 3 = 3 

When x = 1 ⇒ y = 1 + 3 = 4 

When x = 2 ⇒ y = 2 + 3 = 5 

When x = 3 ⇒ y = 3 + 3 = 6 

When x = 4 ⇒ y = 4 + 3 = 7 

When x = 5 y = 5 + 3 = 8 

R = {(0, 3) (1, 4) (2, 5) (3, 6) (4, 7) (5, 8)} 

Domain = {0, 1, 2, 3, 4, 5} 

Range = {3, 4, 5, 6, 7, 8}

Correct answer is red

Correct answer is blue

Correct answer is yellow

Correct answer is fruit

Correct option is B) in

Correct option is A) on

Correct answer is man

Correct answer is pineapple

Answer: 75kmph  

Explanation: First simplify the 7200 second into minutes.  7200/60 = 120 . 

Then simplify into hours.  120/60 = 2 . 

Finally divide the total distance(150km) by the time it took to cover that distance to find the kmph(kilometers per hour).  150/2 = 75  

The final answer is 75kmph.

We have, 

V_rms = \(\frac{V}{\sqrt 2}\) 

∴ V = 220√2 volts.

Voltage difference between A and B,

= 0.2 mA× 5kΩ +0.3 V + 0.2mA × 5kΩ 

= 1 + 0.3 + 1 

= 2.3 V.

(i). In an unbiased p-n junction, 

Holes diffuse from the p- region to n- region because hole concentration in p-region is more as compared to n- region.

(ii). Energy, 

E = \(\frac{hc}{\lambda}\)

= \(\frac{6.6\times10^{-34}\times 3 \times 10^8}{600\times 10{-9}\times 10^{-19}}\)eV

= 2.06 eV

As E< Eg (2.8 eV),

So p-n junction cannot detect the radiation of given wavelength.

n_e = 5 × 10²² – 5 × 10²⁰ 

= (5 - 0.05) × 10²² 

= 4.95 × 10²²m⁻³

∴ n_h = \(\frac{n^2_i}{n_e}\)

= \(\frac{(1.5\times 10^{16})^2}{4.95\times 10^{-22}}\)

= 4.54 × 10⁹m^-3

As n_e>nh, 

So the material is n- type semiconductor.

Getting 3 equations as 

10g + 6f + C = – 34 

2g + 10f + C = -26 

6g – 2f + C = – 10 

Solving to get g = -, f = -2, C = -2 

Final equation: x² + y² – 4x – 4y – 2 = 0

Since, required plane is parallel to the plane 5x - 2y  + z - 2 = 0

\(\therefore\) Normal of given plane is perpendicular to the required plane.

\(\therefore\) Direction ratios of normal of required plane are 5, -2, 1

(\(\because\) Direction ratios of given plane are 5, -2, 1)

\(\because\) Given plane passes through point (2, -1, 3) and having direction ratios of normal 5, -2, 1

\(\therefore\)  Equation of required plane is

5(x - 2) - 2(y + 1) + 1(z - 3) = 0

⇒ 5x - 2y + z - 10 - 2 - 3 = 0

⇒ 5x - 27 + z - 15 = 0

Hence, equation of required plane is 5x - 2y + z - 15 = 0

Let equation of the required circle is 

x² + y² > + 2gy + 2fy + c = 0

Given this equation passes through the points

(1,-4) and (5,2) 

(1,4) (1)² + (-4)² 2g(1) + 2f(-4) + c = 0 

2g - 8f + C +17 = 0 ... (1) 

(5,2) 52 +22 + 2g (5) 2f(2) + c= 0 

10g + 4f + C + 29 = 0 .... (2) 

Also center (-g,-f) lies on the line x - 2 y + 9 = 0 

-g -2(-f) + c = 0 

-g + 2f + c = 0 ... (3) 

Solving equations 1,2 and 3 we get 

G = -3 f = 3,c = -47 

G = 1 f = 2, c = 4 

The required equation are 

x² + y² + 6x - y - 47 = 0 

x² + y² - 2x - 4y + 4 = 0

By hypothesis

1/2|a(b - 2) + 0(2 - 0) + 2 (0 - b)| = 0

⇒ ab -2a - 2b = 0

⇒ 2a + 2b = ab

1/a + 1/b = 1/2