Explore topic-wise InterviewSolutions in Current Affairs.

(1) OR गेट तथा (2) NOT गेट।

क्योंकि इनके निर्गम तथा निवेश के बीच एक तर्कपूर्ण सम्बन्ध होता है।

एक ‘शुद्ध’ अर्द्धचालक, जिसमें कोई अपद्रव्य न मिला हो, ‘निज अर्द्धचालक’ कहलाता है। इस प्रकार, शुद्ध जर्मेनियम तथा सिलिकॉन अपनी प्राकृतिक अवस्था में निज़ अर्द्धचालक हैं। निज अर्द्धचालकों की वैद्युत चालकता अति अल्प होती है। परन्तु यदि किसी ऐसे पदार्थ की बहुत थोड़ी-सी मात्रा, जिसकी संयोजकता 5 अथवा 3 हो, शुद्ध जर्मेनियम (अथवा सिलिकॉन) क्रिस्टल में अपद्रव्य के रूप में मिश्रित कर दें तो क्रिस्टल की चालकता काफी बढ़ जाती है। मिश्रित करने की क्रिया को ‘अपमिश्रण’ । या ‘मादित करना’ कहते हैं। उदाहरणार्थ, 10⁸ जर्मोनियम परमाणुओं में 1 अपद्रव्य परमाणु मिश्रित कर देने पर, जर्मोनियम की चालकता 16 गुना तक बढ़ जाती है। ऐसे अर्द्धचालकों को बाह्य अथवा अपद्रव्य अथवा अपमिश्रित अर्द्धचालक कहते हैं। इन अर्द्धचालकों में मिश्रित किये जाने वाले अपद्रव्य की मात्रा को नियन्त्रित करके इच्छानुसार चालकता अर्जित की जा सकती है।

(iii) ऐलुमिनियम

OR, AND तथा NOT गेट।।

(d) प्रकथन सत्य है।

(iii) p-प्रकार के अर्द्धचालकों में

“प्रतिलोमक द्वार’ क्योंकि यह निवेशी अवस्था का प्रतिलोम कर देता है।

A  +  B  =  Y

चालन बैण्ड तथा संयोजी बैण्ड।

1 निवेश तथा 1 निर्गमा

(i) A = 0, B = 0

किसी निश्चित लघु ऊर्जा परिसर में अत्यन्त निकट रूप से स्थित ऊर्जा स्तरों की एक बड़ी संख्या का समूह ऊर्जा बैण्ड कहलाता है।

(iv) न इलेक्ट्रॉन तथा न कोटर

(iv) इसके दोनों निवेशी 0 हों।

(c) प्रकथन सत्य है।

(b) तथा (c) प्रकथन सत्य हैं।

(iv) A = 1, B = 1

(c) प्रकथन सत्य है।

In △ ABD

Using the Pythagoras theorem

BD² = AB² + AD²

By substituting the values

26² =AB² + 24²

On further calculation

AB² = 676 – 576

By subtraction

AB² = 100

By taking out the square root

AB = √100

So we get

Base = AB = 10cm

We know that area of △ ABD = ½ × b × h

By substituting the values

Area of △ ABD = ½ × 10 × 24

On further calculation

Area of △ ABD = 120 cm²

We know that the area of △ BCD = √3/4 a²

By substituting the values

Area of △ BCD = (1.73/4) (26)²

So we get

Area of △ BCD = 292.37 cm²

So we get area of quadrilateral ABCD = Area of △ ABD + Area of △ BCD

By substituting the values

Area of quadrilateral ABCD = 120 + 29237

By addition

Area of quadrilateral ABCD = 412.37 cm²

The perimeter of quadrilateral ABCD = AB + BC + CD + DA

By substituting the values

Perimeter = 10 + 26 + 26 + 24

So we get

Perimeter = 86cm

Therefore, the area is 412.37 cm² and perimeter is 86cm.

i. 42912

ii. 90000

Correct Answer is Kerala

Given power of lens P = 2D

We know, P = 100/f(in cm) = 2

= 100/f = f = 100/2 = 50 cm

∴ Focal length of lens (f) = 50 cm.

Correct option is  A) 25cm

Correct option is  A) 40 cm

KMnO₄ or any other suitable example.

Scandium (Sc).

Most of the transition metal ions exhibit characteristic colours in aqueous solutions because of d-d transition, as they have maximum number of unpaired electrons.

Bond angle of H₂O is larger, because oxygen is more electronegative than sulphur therefore bond pair electron of O–H bond will be closer to oxygen and there will be more bond-pair bond-pair repulsion between bond pairs of two O–H bonds.

4d and 5d transition elements (2nd and 3rd series) are Iarger in size than the corresponding 3d elements. Hence the valence electrons are less tightly held and form M-M bond more frequently.

(ii), (iii)

(ii) MoO₃

(iii) WO₃

In the crystal lattice, transition elements have interstitial vacant space into which small sized nonmetal atoms such as H,B, C, or N are trapped. These compounds are non-stoichiometric, neither typically ionic nor covalent. e.g., TiC, MH, Fe₃H etc.

(i), (ii)

(i) Cr

(ii) Co

Many of the transition elements are known to form interstitial compounds because of unpaired electrons in the d-orbital. Transition elements have vacant interstitial sites and are able to trap small atoms like H, C or N to form such compound.

(c) C and CO

Cl_(g) + e^- → Cl^- _(g)

∆H = 348 kJ mol^-1 

For one mole (35.5g) 348 kJ is released. 

∴ For 17.5g chlorine, (348 KJ / 35.5 g) x 17.75 energy leased. 

∴ The amount of energy released = 348/2 = 174 kJ.

Other Excretory Organs in Man:

1. Skin: Sweat glands are found in the human skin, which secretes out sweat containing water and some nitrogenous excretory substances.

2. Lungs: Cellular respiration release CO₂ as excretory substance, which is eliminated outside by respiratory process in the lungs. 

3. Liver: Liver cells convert nitrogenous part of amino acids into ammonia and then ammonia into urea. The urea is released into the blood. The liver also forms bile pigments which are excreted along with the bile juice into the intestine.

• It is a hereditary disease in which blood uric acid is increased. 
• It gets deposited in the synovial joints and in Kidney tissues. 
• This disease may cause due to dehydration, fasting and diuresis

Presence of glucose in the urine is called glycosuria.

Bright’s disease or Nephritis: 

• This disease is caused due to infection of Streptococci bacteria in the glomeruli. 
• As a result, they swell and their membrane becomes more permeable to RBC’s and proteins which filtered out and appears in the filtrate. 
• If nephritis is not cured than fluid deposition occurs and the condition of swollen legs happens, which is called Edema or Dropsy.

The process of filtration of blood from the glomerulus into the Bowman’s capsule because of increased glomerular blood pressure is called as Ultrafiltration.

Vasa Recta are ‘U’ shaped, thin walled capillaries that arise from efferent arteriole and run parallel to the Henle’s loop. 

They retain the reabsorbed ions in the medullary tissue fluid and maintain its high osmolarity and are involved in the counter current mechanism.

1. Flame cells

2. Nephridia

The answer is (b) Blood without blood corpuscles and plasma protein

The answer is Urania

Sodium chloride and lactic acid.

The extensions of cortex in between the medullary pyramids as renal columns, are called columns of Bertini.

The answer is Ammonotelic.

Correct answer is (c) Haldane 

Correct answer is (c) Oxygen