Class 12th

π = iCRT
P? = 1 * (10/180) * R * T (For Glucose)
P? = 1 * (10/60) * R * T (For Urea)
P? = 1 * (10/342) * R * T (For Sucrose)

∴ P? > P? > P?

1⁰amines react with Hingsberg's reagent to give a solid, which dissolve in alkali.

Λ°m (NaCl) = 126.45Scm² mol? ¹
Λ°m (HCl) = 426.16Scm² mol? ¹
Λ°m (CH? COONa) = 91Scm² mol? ¹
Λ°m (CH? COOH) =Λ°m (CH? COONa) +Λ°m (HCl)−Λ°m (NaCl)
= 91 + 426.16 – 126.45
= 391.72Scm² mol? ¹

Vitamin B? deficiency → Pernicious anaemia (RBC deficient in heamoglobin)

Fundamental harmonic frequency open pipe

$=\frac{\mathrm{V}}{2\mathrm{L}}={\mathrm{v}}_1$

(say)

Fundamental harmonic frequency of closed pipe $=\frac{\mathrm{V}}{4\mathrm{L}}$

$={\mathrm{v}}_2$ (say)

$\Rightarrow \frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}=\frac{\frac{\mathrm{V}}{2\mathrm{L}}}{\frac{\mathrm{V}}{4\mathrm{L}}}\mathrm{}2:1$

Diameter = main scale reading + (circular scale reading * least count)
Diameter = 0 + (52 * 0.01 mm) = 0.52 mm = 0.052 cm.
In the RLC circuit:
Given I? = 10√2 A, so I? = I? /√2 = 10 A.
V? = √ [V? ² + (V? - V? )²] = √ [40² + (40 - 10)²] = √ [1600 + 30²] = √ [1600 + 900] = √2500 = 50 V.
Impedance Z = V? / I? = 50 V / 10 A = 5 Ω.
For Hindi: I? = 10√2 A, V? = 50V, Z = V? /I? = 50/ (10√2) = 5/√2 Ω.

Yes, candidates can get admission in Sai Tirupati University for BA with 60% in Class 12. The minimum required aggregate for almost all the UG courses is 45%-55% including relaxations for the reserved category. Candidates must ensure that they submit the marksheets and certificate of Class 12 to apply for and confirm admission at Sai Tirupati University.

A reverse-biased Zener diode is used as a voltage regulator.
The potential barrier for Germanium (Ge) is approximately 0.3 V.
The potential barrier for Silicon (Si) is approximately 0.7 V.