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In right shift register, right shift operation of binary $11$ gives

  1. $5.5$
  2. $5$
  3. $6$
  4. $\text{none of these}$
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When we do bit shifting, a bit shifting tells to shift each bit left or right of a number's binary representation. The last bit of the direction is lost, and a 0 bit inserted in the other end.

In right shift register, the bits are shifted in the rightmost direction & a single right shift divides the number (binary) by $2$, two bit right shift divides the number by $4$, three bit right shift divides the number by $8$ , and so on throwing out the remainders.

I'm taking $11_2$ 

So,

$11 >> 1$ gives us $01_2$ (in binary) i.e. $1_{10}$ (in decimal)

And, 

$11 >> 2$ gives us $00_2$ (in binary) i.e. $0_{10}$ (in decimal).

Now, I'm taking $11_{10}\rightarrow $ i.e. $1011_2$ in binary

∴ 

$1011 >> 1$ gives $0101_2$ i.e. $5_{10}$

$1011>>2$ gives $0010_2$ i.e. $2_{10}$

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