NIMCET 2025 Previous Year Questions (PYQs) – Page 2 of 12

• P: ✔ True — DRAM cells use capacitors that leak charge and hence must be refreshed periodically.
• Q: ✔ True — SRAM uses flip-flops, so it retains data without refreshing.
• R: ✗ False — DRAM is actually slower than SRAM.
• S: ✔ True — DRAM is denser and cheaper, making it ideal for main memory.

✅ Correct statements: P, Q, and S

Let’s first convert both binary numbers into decimal (two’s complement form):

• $A = 10011001_2$ → negative because MSB = 1 → invert bits: $01100110_2 = 102_{10}$ → add 1 → $103_{10}$ So, $A = -103$.
• $B = 11010111_2$ → MSB = 1 (negative) → invert bits: $00101000_2 = 40_{10}$ → add 1 → $41_{10}$ So, $B = -41$.

Add them:

$A + B = -103 + (-41) = -144$

Now, in 8-bit two’s complement, the range is $-128$ to $+127$. Since $-144$ is out of range, overflow occurs.

But let’s compute the 8-bit result (ignoring overflow):

$10011001 + 11010111 =$ 10011001
+ 11010111
= 101110000 (9 bits)

Drop the carry beyond 8 bits → 01110000.

Therefore, the resulting 8-bit binary value is:

✅ Result = 01110000₂

(Overflow occurred, actual signed result would have been −144, but the 8-bit stored value is +112.)

• Statement 1 – True: Hardware control units are implemented using combinational logic circuits. They generate control signals directly through hardware, making them faster but difficult to modify since any change requires redesigning the hardware.
• Statement 2 – True: In a horizontal microprogrammed control unit, each control signal is represented by a dedicated bit in the control word. This allows maximum parallelism but results in a wide (long) control word.
• Statement 3 – False: In vertical microprogramming, control words use encoded fields to represent control signals. This makes the word shorter, not longer, but reduces parallelism since only a few signals can be activated at once.
• Statement 4 – True: Microprogrammed control units are easier to implement and modify because control logic is stored in a control memory (microprograms). Modifications can be done by changing microinstructions instead of redesigning hardware.

✅ Correct statements: 1, 2, and 4

Memory size $=512\text{ KB}=2^{19}$ bytes $\Rightarrow$ address field needs $19$ bits.

Registers $=6 \Rightarrow$ bits per register $=\lceil \log_2 6\rceil=3$ bits. For two registers $\Rightarrow 2\times3=6$ bits.

Total operand bits $=19+6=25$.

Opcode bits $=32-25=7 \Rightarrow$ maximum distinct instructions $=2^{7}= {128}$.

✅ Final Matching:
P – 2,  Q – 3,  R – 1,  S – 4

Among the given secondary storage devices, the access time comparison is as follows:

• Magnetic Tape Drive → Very slow (sequential access)
• Optical Drive (CD/DVD) → Slower than HDD (mechanical seek)
• Hard Disk Drive (HDD) → Faster than optical and tape drives, but uses mechanical parts
• Solid State Drive (SSD) → Fastest, as it uses flash memory with no moving parts

✅ Correct Answer: Solid State Drive (SSD)

For an n-bit system:

• 1’s complement range: $-(2^{n-1}-1)$ to $+(2^{n-1}-1)$ ⇒ smallest $=-(2^{8}-1)={-255}$.
• 2’s complement range: $-2^{n-1}$ to $+(2^{n-1}-1)$ ⇒ smallest $=-2^{8}={-256}$.

(9-bit means sign + 8 magnitude bits.)