Explore topic-wise InterviewSolutions in Current Affairs.

1. Capacitor stores and regulates the fluctuation in current and transfers it to the electric circuit. 

2. Capacitor is used to temporarily store electric energy. There are two electric conductors in a capacitor. These are separated by an insulator. Glass, vacuum air, paper or plastics may be used as an insulator in the capacitor. 

3. In an electric connection or circuit, capacitor is used to block direct current in order to allow flow of alternating current.

A. Location based services

LBS is Location based services.

Following are the sources of water pollution - 

• Chemical effluents from industries, agricultural and household waste, are the main sources of water pollution. 

• In many rivers or other water bodies, disposal of industrial & household waste, dumping of garbage, results in water pollution. 

• Oil spills also results in accumulation of oily layers on the water surface which destroys the aquatic ecosystem. 

• Effluents from pesticide & fertilizer industry can cause damage to aquatic ecosystem

B. Worse

Altitude error is always Worse than the horizontal error.

Less than the energy generation from a 16-carbon fatty acid

c) Lidocaine

d) Verapamil 

This drug is associated with Torsades de pointes Sotalol.

d) Quinidine

This drug is used in treating supraventricular tachycardias Digoxin.

b) Trimethaphane

d) Verapamil 

b) Propranolol 

The cell is the structural unit of life with various shapes and size. The size of the cell ranges from 0.5 micrometers to 100 micrometers. Mycoplasma gallicepticum, a parasitic bacterium is thought to be the smallest known organisms with the size from 0.2 to 0.3 micrometer, capable of independent growth and reproduction. Later, the name for mycoplasma was pleuropneumonia-like organisms or PPLO.

Class III agents include: bretylium, amiodarone, ibutilide, sotalol, dofetilide, vernakalant and dronedarone.

Blood group O – has no antigens, but both anti-A and anti-B antibodies in the plasma.

We know the Current formula in L/R circut

I(t) = I_o [1 - e^-t/τ ]

, where I_o = ε/R is the maximum current 

and the time constant τ = L/R 

if current reaches 99 % of steady state,  then 

[1 - e^-t/τ ]

 = 0.99    or e^-t/τ

 = 0.01 = 10^-2  . 

Hence t/τ = 2 × ln10 = 4.6

if current reaches 99.9 % of steady state,  then 

[1 - e^-t/τ ]

 = 0.999 or e^-t/τ 

 = 0.001 = 10^-3  . 

Hence t/τ = 3 × ln10 = 6.9

Embryogenesis, the first eight weeks of development after fertilization, is an incredibly complicated process. It’s amazing that in eight weeks we’re transforming from a single cell to an organism with a multi-level body plan.

Condenser back- pressure must be low, because steam should be dumped into the condenser so as to recycle it to boiler through the recycle process. It improves efficiency of the turbine, as the heat rejection is less. It is achieved by the help of ejectors and also passing cold water in the condenser through the tubes of the condenser so that maximum vacuum can be obtained.  

Addressing modes of 8086 :

When 8086 executes an instruction, it performs the specified function on data. These data are called its operands and may be part of the instruction, reside in one of the internal registers of the microprocessor, stored at an address in memory or held at an I/O port, to access these different types of operands, the 8086 is provided with various addressing modes (Data Addressing Modes).

Data Addressing Modes of 8086 

The 8086 has 12 addressing modes. The various 8086 addressing modes can be classified into five groups. 

A. Addressing modes for accessing immediate and register data (register and immediate modes). 

B. Addressing modes for accessing data in memory (memory modes) 

C. Addressing modes for accessing I/O ports (I/O modes) 

D. Relative addressing mode 

E. Implied addressing mode 

8086 ADDRESSING MODES

A. Immediate addressing mode: 

In this mode, 8 or 16 bit data can be specified as part of the instruction. OP Code Immediate Operand 

Example 1 : MOV CL, 03 H 

Moves the 8 bit data 03 H into CL 

Example 2 : MOV DX, 0525 H 

Moves the 16 bit data 0525 H into DX In the above two examples, the source operand is in immediate mode and the destination operand is in register mode. 

A constant such as “VALUE” can be defined by the assembler EQUATE directive such as VALUE EQU 35H

Example : MOV BH, VALUE 

Used to load 35 H into BH 

Register addressing mode : 

The operand to be accessed is specified as residing in an internal register of 8086. Fig. below shows internal registers, any one can be used as a source or destination operand, however only the data registers can be accessed as either a byte or word. 

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Register Operand sizes 

Byte (Reg 8) Word (Reg 16) 

Accumulator AL, AH Ax 

Base BL, BH Bx 

Count CL, CH Cx 

Data DL, DH Dx 

Stack pointer - SP 

Base pointer - BP 

Source index - SI 

Destination index - DI 

Code Segment - CS 

Data Segment - DS 

Stack Segment - SS 

Extra Segment - ES 

Example 1 : MOV DX (Destination Register) , CX (Source Register) Which moves 16 bit content of CS into DX. 

Example 2 : MOV CL, DL

Moves 8 bit contents of DL into CL 

MOV BX, CH is an illegal instruction.

The register sizes must be the same. 

B. Direct addressing mode : 

The instruction Opcode is followed by an affective address, this effective address is directly used as the 16 bit offset of the storage location of the operand from the location specified by the current value in the selected segment register. 

The default segment is always DS. 

The 20 bit physical address of the operand in memory is normally obtained as 

PA = DS : EA 

But by using a segment override prefix (SOP) in the instruction, any of the four segment registers can be referenced, 

PA = CS 

DS : Direct Address 

SS 

ES

The Execution Unit (EU) has direct access to all registers and data for register and immediate operands. However the EU cannot directly access the memory operands. It must use the BIU, in order to access memory operands.

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In the direct addressing mode, the 16 bit effective address (EA) is taken directly from the displacement field of the instruction.

Example 1 :MOV CX, START 

If the 16 bit value assigned to the offset START by the programmer using an assembler pseudo instruction such as DW is 0040 and [DS] = 3050. Then BIU generates the 20 bit physical address 30540 H. 

The content of 30540 is moved to CL 

The content of 30541 is moved to CH 

Example 2 :MOV CH, START

If [DS] = 3050 and START = 0040

8 bit content of memory location 30540 is moved to CH.

Example 3 :MOV START, BX 

With [DS] = 3050, the value of START is 0040. 

Physical address : 30540 

MOV instruction moves (BL) and (BH) to locations 30540 and 30541 respectively. 

Register indirect addressing mode : 

The EA is specified in either pointer (BX) register or an index (SI or DI) register. The 20 bit physical address is computed using DS and EA. 

Example : MOV [DI], BX 

register indirect

If [DS] = 5004, [DI] = 0020, [Bx] = 2456 PA=50060. 

The content of BX(2456) is moved to memory locations 50060 H and 50061 H.

CS 

PA = DS BX 

SS = SI 

ES DI

Based addressing mode:

CS

PA = DS BX 

SS : or + displacement 

ES BP

when memory is accessed PA is computed from BX and DS when the stack is accessed PA is computed from BP and SS.

Example : MOV AL, START [BX]

or MOV AL, [START + BX] 

based mode

EA : [START] + [BX]

PA : [DS] + [EA] 

The 8 bit content of this memory location is moved to AL. 

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Indexed addressing mode:

CS

PA = DS SI

SS : or + 8 or 16bit displacement

ES DI

Example : MOV BH, START [SI]

PA : [SART] + [SI] + [DS] 

The content of this memory is moved into BH. 

Based Indexed addressing mode:

CS

PA = DS BX SI

SS : or + or + 8 or 16 bit displacement

ES BP DI

Example : MOV ALPHA [SI] [BX], CL

If [BX] = 0200, ALPHA – 08, [SI] = 1000 H and [DS] = 3000

Physical address (PA) = 31208

8 bit content of CL is moved to 31208 memory address. 

String addressing mode: 

The string instructions automatically assume SI to point to the first byte or word of the source operand and DI to point to the first byte or word of the destination operand. The contents of SI and DI are automatically incremented (by clearing DF to 0 by CLD instruction) to point to the next byte or word. 

Example : MOV S BYTE 

If [DF] = 0, [DS] = 2000 H, [SI] = 0500,

[ES] = 4000, [DI] = 0300

Source address : 20500, assume it contains 38 

PA : [DS] + [SI]

Destination address : [ES] + [DI] = 40300, assume it contains 45 

After executing MOV S BYTE,

[40300] = 38

[SI] = 0501 incremented 

[DI] = 0301

C. I/O mode (direct) :

Port number is an 8 bit immediate operand. 

Example : OUT 05 H, AL 

Outputs [AL] to 8 bit port 05 H

I/O mode (indirect): 

The port number is taken from DX.

Example 1 : INAL, DX 

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OR

OR

OR

If [DX] = 5040

8 bit content by port 5040 is moved into AL. 

Example 2 : IN AX, DX 

Inputs 8 bit content of ports 5040 and 5041 into AL and AH respectively. 

D. Relative addressing mode: 

Example : JNC START 

If CY=O, then PC is loaded with current PC contents plus 8 bit signed value of START, otherwise the next instruction is executed. 

E. Implied addressing mode:

Instruction using this mode have no operands. 

Example : CLC which clears carry flag to zero. 

SINGLE INDEX DOUBLE INDEX 

Fig.3.1 : Summary of 8086 

Addressing Modes Encoded in the instruction

BX 

OR 

BP 

SI 

OR 

DI

+ 

+ + 

+ + 

CS 0000 

PHYSICAL ADDRESS 

DS 0000 

SS 0000 

ES 0000 

DISPLACEMENT 

Explicit in the 

instruction 

Assumed 

unless 

over 

ridden 

by prefix 

EU BIU 

BX 

OR 

BP 

OR 

SI

OR 

DI 

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(a) \(sin^{-1}\left(\frac{6x}{1+9x^2}\right)=sin^{-1}\left(\frac{2(3x)}{1+(3x)^2}\right)\)

Let 3x \(=tan\theta\)

\(\Rightarrow\theta=tan^{-1}(3x)\)

Then, \(sin^{-1}\left(\frac{2\times3x}{1+(3x)^2}\right)=sin^{-1}\left(\frac{2tan\theta}{1+tan^2\theta}\right)\)

\(=sin^{-1}(sin\,2\theta)\) \((\because sin\,2\theta=\frac{2tan\theta}{1+tan^2\theta})\)

\(=2\theta\)

\(=2\,tan^{-1}3x\,\,(\because\theta=tan^{-1}3x)\) 

 \(=2tan^{-1}3x=2tan^{-1}(ax)\)

\(\therefore\) a = 3

 Correct answer is a. microorganisms

 Correct answer is c. Personality

 Correct answer is c. Early age

 Correct answer is b. Personal hygiene

 Correct answer is c. Ergonomics

Correct option: c) Blanching

Enoki is a type of Mushrooms.

Correct option: a) Caramelisation